Electrostatics
A point charge $q=1 \mu \mathrm{C}$ is located at a distance 2 cm from one end of a thin insulating wire of length 10 cm having a charge $Q=24 \mu \mathrm{C}$, distributed uniformly along its length, as shown in figure. Force between $q$ and wire is $\_\_\_\_$ N.
(Use : $\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{~N} \cdot \mathrm{~m}^2 / \mathrm{C}^2$ )
Explanation:
Because the charge on the wire is distributed, we consider a small element dx on the wire at a distance x from the point charge q.
Linear charge density of the line charge is $\lambda=\frac{\mathrm{Q}}{\mathrm{L}}$
Charge on element $\mathrm{dQ}=\lambda \mathrm{dx}$
Small force acting between the charges is,
$ \mathrm{dF}=\frac{1}{4 \pi \epsilon_{\mathrm{o}}} \frac{\mathrm{q} \cdot \mathrm{dQ}}{\mathrm{x}^2}=\frac{\mathrm{kq} \mathrm{q} \lambda \mathrm{dx}}{\mathrm{x}^2} $
To find the total force F , we integrate from the nearest end $(\mathrm{x}=2 \mathrm{~cm}=0.02 \mathrm{~m})$ to the farthest end ( $\mathrm{x}=2 \mathrm{~cm}+10 \mathrm{~m}=12 \mathrm{~cm}=0.12 \mathrm{~m}$ ):
$ \mathrm{F}=\int\limits_{0.02}^{0.12} \frac{\mathrm{kq} \lambda}{\mathrm{x}^2} \mathrm{dx}=\mathrm{kq} \lambda\left[-\frac{1}{\mathrm{x}}\right]_{0.02}^{0.12} $
$\Rightarrow $ $\mathrm{F}=\mathrm{kq} \lambda\left(\frac{1}{0.02}-\frac{1}{0.12}\right)=\mathrm{kq} \lambda\left(\frac{6-1}{0.12}\right)=\frac{5 \mathrm{kq} \lambda}{0.12}$
$\Rightarrow $ $\mathrm{F}=\frac{5 \mathrm{kq}}{0.12} \times \frac{\mathrm{Q}}{\mathrm{L}}$
Substituting the given values,
$\mathrm{k}=9 \times 10^9 ; \mathrm{q}=10^{-6} ; \mathrm{Q}= 24 \times 10^{-6} ; \mathrm{L}=10 \mathrm{~cm}=0.10 \mathrm{~m} $
$ \mathrm{~F}=\frac{5 \times 9 \times 10^9 \times 1 \times 10^{-6} \times 24 \times 10^{-6}}{0.10 \times 0.12} \mathrm{~N} $
$\begin{aligned} \Rightarrow \mathrm{F} & =\frac{216}{2.4} \\ \Rightarrow \mathrm{~F} & =90 \mathrm{~N}\end{aligned}$
Therefore, the electrostatic force exerted by the finite wire on the point charge is 90 N .
Hence, the correct answer is 90 .
A three coulomb charge moves from the point $(0,-2,-5)$ to the point $(5,1,2)$ in an electric field expressed as $\vec{E}=2 x \hat{\mathrm{i}}+3 \mathrm{y}^2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \mathrm{N} / \mathrm{C}$. The work done in moving the charge is $\_\_\_\_$ J.
Explanation:
The work done (W) by an external agent in moving a charge (q) in an electric field ( $\overrightarrow{\mathrm{E}}$ ) is given by the change in potential energy, which is the negative of the work done by the field.
The electric force acting on the charge by electric field is $\overrightarrow{\mathrm{F}}=\mathrm{q} \overrightarrow{\mathrm{E}}$
So, the work done is :
$ \mathrm{W}=\mathrm{q} \int \overrightarrow{\mathrm{E}} \cdot \mathrm{~d} \overrightarrow{\mathrm{r}} $
Where $\mathrm{d} \overrightarrow{\mathrm{r}}=\mathrm{dxî}+\mathrm{dy} \hat{\mathrm{j}}+\mathrm{dz} \hat{\mathrm{k}}$ is the infinitesimal displacement vector.
$ \mathrm{W}=\mathrm{q} \int\limits_{(0,-2,-5)}^{(5,1,2)}\left(2 \mathrm{x} \hat{\imath}+3 \mathrm{y}^2 \hat{\jmath}+4 \hat{\mathrm{k}}\right) \cdot(\mathrm{dx} \hat{\imath}+\mathrm{dy} \hat{\jmath}+\mathrm{dz} \hat{\mathrm{k}}) $
$\Rightarrow $ $\mathrm{W}=\mathrm{q}\left[\int_0^5 2 \mathrm{xdx}+\int_{-2}^1 3 \mathrm{y}^2 \mathrm{dy}+\int_{-5}^2 4 \mathrm{dz}\right]$
$\Rightarrow $ $\mathrm{W}=\mathrm{q}\left[2\left[\frac{\mathrm{x}^2}{2}\right]_0^5+3\left[\frac{\mathrm{y}^3}{3}\right]_{-2}^1+4[\mathrm{x}]_{-5}^2\right]$
$\Rightarrow $ $\mathrm{W}=3\left[\left[5^2-0^2\right]+\left[1^3-(-2)^3\right]+4[2-(-5)]\right]$
$\Rightarrow $ $\mathrm{W}=3 \times(25+9+28) $
$\Rightarrow $ $ \mathrm{W}=3 \times 62=186 \mathrm{~J}$
Therefore, the work done in moving the charge is 186 J.
Explanation:
To calculate the area of the surface through which the electric flux is determined, we use the given electric field and the specified surface orientation. The electric field is expressed as $\overrightarrow{\mathrm{E}} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^3 \, \mathrm{N/C}$.
Since the surface is parallel to the $x-z$ plane, its area vector $\overrightarrow{\mathrm{A}}$ is directed along the $y$-axis, making it $\mathrm{~A} \hat{\mathrm{j}}$.
The electric flux $\phi$ through the surface is given by:
$ \phi = \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{~A}} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^3 \cdot \mathrm{~A} \hat{\mathrm{j}} $
Calculating the dot product, the only component contributing to the flux is the $y$-component:
$ \phi = (4 \times 10^3) \mathrm{~A} $
Given that the electric flux $\phi$ is $6.0 \, \mathrm{Nm}^2/\mathrm{C}$, we can equate and solve for $\mathrm{A}$:
$ 6 = 4 \times 10^3 \mathrm{~A} $
$ \mathrm{A} = \frac{6}{4 \times 10^3} = 1.5 \times 10^{-3} \, \mathrm{m}^2 $
Converting this area from square meters to square centimeters:
$ \mathrm{A} = 1.5 \times 10^{-3} \, \mathrm{m}^2 = 15 \, \mathrm{cm}^2 $
Explanation:
Given values:
The dipole moment is $ p = 6 \times 10^{-6} \ \mathrm{Cm} $.
The electric field is $ E = 10^6 \ \mathrm{V/m} $.
Formula for work done:
Work required to rotate the dipole is $ W = \Delta U = -pE(\cos \theta_f - \cos \theta_i) $ where $ \theta_i $ is the initial angle between the dipole and the electric field, and $ \theta_f $ is the final angle.
Angles:
At the start, the dipole is parallel to the field ($ \theta_i = 0^\circ $), so $\cos \theta_i = 1$.
At the end, the dipole is opposite the field ($ \theta_f = 180^\circ $), so $\cos \theta_f = -1$.
Calculation:
$ W = -pE (\cos 180^\circ - \cos 0^\circ) = -pE(-1 - 1) = 2pE $
Plug in the values:
$ W = 2 \times (6 \times 10^{-6}) \times (10^6) = 2 \times 6 = 12\ \mathrm{J} $
A square loop of sides $a=1 \mathrm{~m}$ is held normally in front of a point charge $\mathrm{q}=1 \mathrm{C}$ at a distance $\frac{\mathrm{a}}{2}$. The flux of the electric field through the shaded region is $\frac{5}{\mathrm{p}} \times \frac{1}{\varepsilon_0} \frac{\mathrm{Nm}^2}{\mathrm{C}}$, where the value of p is ________ .
Explanation:
Total flux through square $=\frac{\mathrm{q}}{\epsilon_0}\left(\frac{1}{6}\right)$
Lets divide square is 8 equal parts.
Flux is same for each part.
$\therefore$ Flux through shaded portion is $\frac{5}{8}$ (Total flux)
$ =\frac{5}{8} \times \frac{\mathrm{q}}{\epsilon_0} \frac{1}{6}=\frac{5}{48} \frac{1}{\epsilon_0} $
$\therefore$ Required Ans. is 48
A positive ion $A$ and a negative ion $B$ has charges $6.67 \times 10^{-19} \mathrm{C}$ and $9.6 \times 10^{-10} \mathrm{C}$, and masses $19.2 \times 10^{-27} \mathrm{~kg}$ and $9 \times 10^{-27} \mathrm{~kg}$ respectively. At an instant, the ions are separated by a certain distance $r$. At that instant the ratio of the magnitudes of electrostatic force to gravitational force is $\mathrm{P} \times 10^{-13}$, where the value of P is _________.
(Take $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-1}$ and universal gravitational constant as $6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}$ )
Explanation:
To find the ratio of the magnitudes of electrostatic force to gravitational force between ions $A$ and $B$, we use the following formulas for electrostatic force ($F_e$) and gravitational force ($F_g$):
$ F_e = \frac{k \cdot q_1 \cdot q_2}{r^2} $
$ F_g = \frac{G \cdot m_1 \cdot m_2}{r^2} $
To find the ratio $\frac{F_e}{F_g}$, simplify as follows:
$ \frac{F_e}{F_g} = \frac{k \cdot q_1 \cdot q_2}{G \cdot m_1 \cdot m_2} $
Using the given values:
$k = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-1}$
$q_1 = 6.67 \times 10^{-19} \, \text{C}$
$q_2 = 9.6 \times 10^{-10} \, \text{C}$
$G = 6.67 \times 10^{-11} \, \text{Nm}^2 \text{kg}^{-2}$
$m_1 = 19.2 \times 10^{-27} \, \text{kg}$
$m_2 = 9 \times 10^{-27} \, \text{kg}$
The formula becomes:
$ \frac{F_e}{F_g} = \frac{9 \times 10^9 \times 6.67 \times 10^{-19} \times 9.6 \times 10^{-10}}{6.67 \times 10^{-11} \times 19.2 \times 10^{-27} \times 9 \times 10^{-27}} $
Calculate the result:
$ = \frac{10^{-20}}{2 \times 10^{-65}} $
This calculation gives the ratio of the electrostatic force to the gravitational force without considering their separation distance $r$, as it cancels out. The value of $P$ mentioned in the prompt can be deduced from the simplified final expression.
An electric field $\vec{E}=(2 x \hat{i}) N C^{-1}$ exists in space. A cube of side $2 \mathrm{~m}$ is placed in the space as per figure given below. The electric flux through the cube is ______ $\mathrm{Nm}^2 / \mathrm{C}$.
Explanation:
Flux will only be due to surfaces having area vector parallel to $x$ - axis
$\begin{aligned} \therefore \quad & \phi_{\text {net }}=A[8-4] \\ & =4 A=4 \times 4=16 \end{aligned}$
At the centre of a half ring of radius $\mathrm{R}=10 \mathrm{~cm}$ and linear charge density $4 \mathrm{~nC} \mathrm{~m}^{-1}$, the potential is $x \pi \mathrm{V}$. The value of $x$ is _________.
Explanation:
$\begin{aligned} V & =\frac{K Q}{R} \\ & =\frac{9 \times 10^9 \times 4 \times 10^{-9} \pi R}{R} \\ & =36 \pi \end{aligned}$
If the net electric field at point $\mathrm{P}$ along $\mathrm{Y}$ axis is zero, then the ratio of $\left|\frac{q_2}{q_3}\right|$ is $\frac{8}{5 \sqrt{x}}$, where $x=$ ________.
Explanation:
$\begin{aligned} \Rightarrow \quad & E_2 \cos \theta=E_3 \cos \phi \\ & \frac{q_2}{\left(2^2+4^2\right)} \frac{4}{\sqrt{2^2+4^2}}=\frac{q_3}{4^2+3^2} \frac{4}{\sqrt{4^2+3^2}} \\ & \frac{q_2}{q_3}=\frac{20}{25} \frac{\sqrt{20}}{5}=\frac{4}{5} \times \frac{2 \sqrt{5}}{5} \\ \Rightarrow \quad & n=5 \end{aligned}$
An electric field, $\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}$ passes through the surface of $4 \mathrm{~m}^2$ area having unit vector $\hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$. The electric flux for that surface is _________ $\mathrm{Vm}$.
Explanation:
The electric flux through a surface is given by the formula:
$\Phi = \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} = |\overrightarrow{\mathrm{E}}||\overrightarrow{\mathrm{A}}|\cos\theta$
where $\overrightarrow{\mathrm{E}}$ is the electric field, $\overrightarrow{\mathrm{A}}$ is the area vector (with magnitude equal to the area of the surface and direction perpendicular to the surface, defined by the unit vector $\hat{n}$), and $\theta$ is the angle between $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{A}}$. However, when using unit vectors to describe the directions of $\overrightarrow{\mathrm{E}}$ and $\hat{n}$, the dot product can be used to simplify the calculation as follows:
$\Phi = \overrightarrow{\mathrm{E}} \cdot \left(\overrightarrow{\mathrm{A}}\right) = (\overrightarrow{\mathrm{E}} \cdot \hat{n})A$
Given that $\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}$ and $\hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$, and the area $A = 4 \mathrm{m}^2$, we can substitute them into our formula. Note that since $\overrightarrow{\mathrm{A}} = A\hat{n}$, the magnitude of the area vector is the area of the surface itself. First, let's find $\overrightarrow{\mathrm{E}} \cdot \hat{n}$:
$\overrightarrow{\mathrm{E}} \cdot \hat{n} = \left(\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}\right) \cdot \left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$
To compute the dot product, we multiply corresponding components and then add them up:
$\left(\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}\right) \cdot \left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right) = \frac{1}{6}(2\cdot2 + 6\cdot1 + 8\cdot1)$
$= \frac{1}{6}(4 + 6 + 8) = \frac{18}{6} = 3$
Then, the electric flux through the surface is:
$\Phi = (\overrightarrow{\mathrm{E}} \cdot \hat{n})A = 3 \times 4 \mathrm{m}^2$
$\Phi = 12 \mathrm{Vm}$
So, the electric flux for that surface is $12 \mathrm{Vm}$.
Three infinitely long charged thin sheets are placed as shown in figure. The magnitude of electric field at the point $P$ is $\frac{x \sigma}{\epsilon_0}$. The value of $x$ is _________ (all quantities are measured in SI units).
Explanation:
$\begin{aligned} \overrightarrow{\mathrm{E}}_{\mathrm{p}} & =\left(\frac{\sigma}{2 \varepsilon_0}+\frac{2 \sigma}{2 \varepsilon_0}+\frac{\sigma}{2 \varepsilon_0}\right)(-\hat{\mathrm{i}}) \\\\ & =-\frac{2 \sigma}{\varepsilon_0} \hat{\mathrm{i}}\end{aligned}$
The electric field at point $\mathrm{p}$ due to an electric dipole is $\mathrm{E}$. The electric field at point $\mathrm{R}$ on equitorial line will be $\frac{\mathrm{E}}{x}$. The value of $x$ :
Explanation:
$\begin{aligned} & E=\frac{2 k p}{r^3} \\ & E_R=\frac{k p}{(2 r)^3}=\frac{1}{8}\left(\frac{E}{2}\right) \\ & =\frac{E}{16} \\ & \therefore \quad x=16 \\ & \end{aligned}$
An infinite plane sheet of charge having uniform surface charge density $+\sigma_{\mathrm{s}} \mathrm{C} / \mathrm{m}^2$ is placed on $x$-$y$ plane. Another infinitely long line charge having uniform linear charge density $+\lambda_e \mathrm{C} / \mathrm{m}$ is placed at $z=4 \mathrm{~m}$ plane and parallel to $y$-axis. If the magnitude values $\left|\sigma_{\mathrm{s}}\right|=2\left|\lambda_{\mathrm{e}}\right|$ then at point $(0,0,2)$, the ratio of magnitudes of electric field values due to sheet charge to that of line charge is $\pi \sqrt{n}: 1$. The value of $n$ is _________.
Explanation:
Given $\sigma_s=2 \lambda_e$
At point $P, E_S=\frac{\sigma_S}{2 \varepsilon_0}$
$\begin{aligned} & E_I=\frac{\lambda_e}{2 \pi r \varepsilon_0} \\ & \frac{E_S}{E_I}=4 \pi: 1=\pi \sqrt{n}: 1 \end{aligned}$
For value of $n=16$
Then the charge on the particle will be $\frac{1}{\sqrt{x}} \mu \mathrm{C}$ where $x=$ ___________ . [use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ]
Explanation:
$\begin{aligned} & \sin \theta=\frac{10}{20}=\frac{1}{2} \\\\ & \theta=30^{\circ} \\\\ & \tan \theta=\frac{\mathrm{qE}}{\mathrm{mg}} \\\\ & \tan 30^{\circ}=\frac{\mathrm{q} \times 2 \times 10^4}{1 \times 10^{-3} \times 10}\end{aligned}$
$\begin{aligned} & \frac{1}{\sqrt{3}}=q \times 10^6 \\\\ & q=\frac{1}{\sqrt{3}} \times 10^{-6} C \\\\ & x=3\end{aligned}$
(Take density of water $=1 \mathrm{~g} / \mathrm{cc}$ )
Explanation:
In air $\tan \frac{\theta}{2}=\frac{F}{m g}=\frac{q^2}{4 \pi \varepsilon_0 r^2 m g}$
In water $\tan \frac{\theta}{2}=\frac{\mathrm{F}^{\prime}}{\mathrm{mg}^{\prime}}=\frac{\mathrm{q}^2}{4 \pi \varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{r}^2 \mathrm{mg}_{\text {eff }}}$
Equate both equations
$ \begin{aligned} & \varepsilon_0 g=\varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{g}\left[1-\frac{1}{1.5}\right] \\\\ & \varepsilon_{\mathrm{r}}=3 \end{aligned} $
The distance between charges $+q$ and $-q$ is $2 l$ and between $+2 q$ and $-2 q$ is $4 l$. The electrostatic potential at point $P$ at a distance $r$ from center $O$ is $-\alpha\left[\frac{q l}{r^2}\right] \times 10^9 \mathrm{~V}$, where the value of $\alpha$ is __________. (Use $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~Nm}^2 \mathrm{C}^{-2}$)
Explanation:
$\begin{aligned} & \mathrm{V}=\frac{\mathrm{K} \overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}}{\mathrm{r}^3}=\frac{9 \times 10^9(6 \mathrm{q} \ell)}{\mathrm{r}^2} \cos \left(120^{\circ}\right) \\\\ & =-(27)\left(\frac{\mathrm{q} \ell}{\mathrm{r}^2}\right) \times 10^9 \mathrm{Nm}^2 \mathrm{c}^{-2} \\\\ & \Rightarrow \alpha=27 \end{aligned}$
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $37^{\circ}$ with each other. When suspended in a liquid of density $0.7 \mathrm{~g} / \mathrm{cm}^3$, the angle remains same. If density of material of the sphere is $1.4 \mathrm{~g} / \mathrm{cm}^3$, the dielectric constant of the liquid is _______ $\left(\tan 37^{\circ}=\frac{3}{4}\right)$
Explanation:
$\begin{aligned} & T \cos \theta=m g \\ & T \sin \theta=F_e \\ & \tan \theta=\frac{F_e}{m g} \end{aligned}$
$\tan \theta=\frac{F_c}{\rho_B V g}$ ..... (i)
$\tan \theta=\frac{F_e}{\frac{k}{\left(\rho_B-\rho_L\right) V g}}$ ..... (ii)
From Eq. (i) & (ii)
$\begin{aligned} & \rho_{\mathrm{B}} \mathrm{Vg}=\left(\rho_{\mathrm{B}}-\rho_{\mathrm{L}}\right) \mathrm{kVg} \\ & 1.4=0.7 \mathrm{k} \\ & \mathrm{k}=2 \end{aligned}$
An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet $\mathrm{S}$ having surface charge density $+\sigma$. The electron at $t=0$ is at a distance of $1 \mathrm{~m}$ from $S$ and has a speed of $1 \mathrm{~m} / \mathrm{s}$. The maximum value of $\sigma$ if the electron strikes $S$ at $t=1 \mathrm{~s}$ is $\alpha\left[\frac{m \epsilon_0}{e}\right] \frac{C}{m^2}$, the value of $\alpha$ is ___________.
Explanation:
$\begin{aligned} & \mathrm{u}=1 \mathrm{~m} / \mathrm{s} ; \mathrm{a}=-\frac{\sigma \mathrm{e}}{2 \varepsilon_0 \mathrm{~m}} \\ & \mathrm{t}=1 \mathrm{~s} \\ & \mathrm{~S}=-1 \mathrm{~m} \\ & \text { Using } \mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 \\ & -1=1 \times 1-\frac{1}{2} \times \frac{\sigma \mathrm{e}}{2 \varepsilon_0 \mathrm{~m}} \times(1)^2 \\ & \therefore \sigma=8 \frac{\varepsilon_0 \mathrm{~m}}{\mathrm{e}} \\ & \therefore \alpha=8 \end{aligned}$
Two charges of $-4 \mu \mathrm{C}$ and $+4 \mu \mathrm{C}$ are placed at the points $\mathrm{A}(1,0,4) \mathrm{m}$ and $\mathrm{B}(2,-1,5) \mathrm{m}$ located in an electric field $\overrightarrow{\mathrm{E}}=0.20 \hat{i} \mathrm{~V} / \mathrm{cm}$. The magnitude of the torque acting on the dipole is $8 \sqrt{\alpha} \times 10^{-5} \mathrm{Nm}$, where $\alpha=$ _________.
Explanation:
$\begin{aligned} & \vec{\tau}=\vec{p} \times \vec{E} \\ & \vec{p}=q \vec{\ell} \\ & \overrightarrow{\mathrm{E}}=0.2 \frac{\mathrm{V}}{\mathrm{cm}}=20 \frac{\mathrm{V}}{\mathrm{m}} \\ & \overrightarrow{\mathrm{p}}=4 \times(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & =(4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \mu \mathrm{C}-\mathrm{m} \\ & \vec{\tau}=(4 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times(20 \hat{\mathrm{i}}) \times 10^{-6} \mathrm{Nm} \\ & =(8 \hat{\mathrm{k}}+8 \hat{\mathrm{j}}) \times 10^{-5}=8 \sqrt{2} \times 10^{-5} \\ & \alpha=2 \end{aligned}$
The electric potential at the surface of an atomic nucleus $(z=50)$ of radius $9 \times 10^{-13} \mathrm{~cm}$ is __________ $\times 10^6 \mathrm{~V}$.
Explanation:
$\begin{aligned} & \text { Potential }=\frac{\mathrm{kQ}}{\mathrm{R}}=\frac{\mathrm{k} . \mathrm{Ze}}{\mathrm{R}} \\ & =\frac{9 \times 10^9 \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-13} \times 10^{-2}} \\ & =8 \times 10^6 \mathrm{~V} \end{aligned}$
A thin metallic wire having cross sectional area of $10^{-4} \mathrm{~m}^2$ is used to make a ring of radius $30 \mathrm{~cm}$. A positive charge of $2 \pi \mathrm{~C}$ is uniformly distributed over the ring, while another positive charge of 30 $\mathrm{pC}$ is kept at the centre of the ring. The tension in the ring is ______ $\mathrm{N}$; provided that the ring does not get deformed (neglect the influence of gravity). (given, $\frac{1}{4 \pi \epsilon_0}=9 \times 10^9$ SI units)
Explanation:
$\begin{aligned} & 2 \mathrm{~T} \sin \frac{\mathrm{d} \theta}{2}=\frac{\mathrm{kq}_0}{\mathrm{R}^2} \cdot \lambda \mathrm{Rd} \theta \\\\ & {\left[\lambda=\frac{\mathrm{Q}}{2 \pi \mathrm{R}}\right]} \end{aligned}$
$\begin{aligned} & \Rightarrow \mathrm{T}=\frac{\mathrm{Kq}_0 \mathrm{Q}}{\left(\mathrm{R}^2\right) \times 2 \pi} \\\\ & =\frac{\left(9 \times 10^9\right)\left(2 \pi \times 30 \times 10^{-12}\right)}{(0.30)^2 \times 2 \pi} \\\\ & =\frac{9 \times 10^{-3} \times 30}{9 \times 10^{-2}}=3 \mathrm{~N} \end{aligned}$
A charge is kept at the central point $\mathrm{P}$ of a cylindrical region. The two edges subtend a half-angle $\theta$ at $\mathrm{P}$, as shown in the figure. When $\theta=30^{\circ}$, then the electric flux through the curved surface of the cylinder is $\Phi$. If $\theta=60^{\circ}$, then the electric flux through the curved surface becomes $\Phi / \sqrt{n}$, where the value of $n$ is ___.
Explanation:
Place a point charge $q$ at the centre $P$ of a cylindrical Gaussian surface. Because of symmetry, the electric flux across each circular end cap is the same; call this $\phi_f$. Let $\phi$ be the flux through the curved side. Applying Gauss’s law to the closed cylinder, the total flux is
$ \phi_{\text{total}} \;=\; 2\phi_f \;+\; \phi \;=\; \frac{q}{\varepsilon_0}. $
The solid angle subtended at point $P$ by the flat surface (spherical cap of half-angle $\theta$ ) is $\Omega_f=2 \pi(1-\cos \theta$ ). The total solid angle is $\Omega_t=4 \pi$. Thus, the flux through one flat surface is
$ \phi_f=\phi_t\left(\frac{\Omega_f}{\Omega_t}\right)=\frac{\phi_t(1-\cos \theta)}{2} $
Solve the above equations to get
$ \phi=\frac{q \cos \theta}{\epsilon_0}=\frac{q \cos 30^{\circ}}{\epsilon_0}=\frac{\sqrt{3} q}{2 \epsilon_0} $
The flux through the curved surface for $\theta=60^{\circ}$ is
$ \phi^{\prime}=\frac{q \cos 60^{\circ}}{\epsilon_0}=\frac{q}{2 \epsilon_0}=\frac{\phi}{\sqrt{3}} $
An infinitely long thin wire, having a uniform charge density per unit length of $5 \mathrm{nC} / \mathrm{m}$, is passing through a spherical shell of radius $1 \mathrm{~m}$, as shown in the figure. A $10 \mathrm{nC}$ charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points $\mathrm{P}$ and $\mathrm{R}$, in Volt, is ________.
[Given: In SI units $\frac{1}{4 \pi \epsilon_0}=9 \times 10^9, \ln 2=0.7$. Ignore the area pierced by the wire.]
Explanation:
due to wire
$\begin{aligned} & \mathrm{dV}=-\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dx}} \\ & \int_\limits{\mathrm{v}_{\mathrm{P}}}^{\mathrm{v}_{\mathrm{R}}} \mathrm{dV}=-\int_\limits{0.5}^2 \frac{2 \mathrm{k} \lambda}{\mathrm{x}} \mathrm{dx} \\ & \mathrm{v}_{\mathrm{R}}-\mathrm{v}_{\mathrm{P}}=-2 \mathrm{k} \lambda \ln \frac{2}{0.5} \\ & =-2 \times 9 \times 10^9 \times 3 \times 10^{-9} \times 2 \times 0.7=-126 \mathrm{~V} \end{aligned}$
due to sphere
$\begin{aligned} & v_R-v_P= \frac{k Q}{2}-\frac{k Q}{1}=-\frac{k Q}{2}=\frac{-9 \times 10^9 \times 10 \times 10^{-9}}{2} \\ &=-45 \mathrm{~V} \\ & v_R-v_P=-126-45=-171 \mathrm{~V} \\ & v_P-v_R= 171 \mathrm{~V} \end{aligned}$
Three point charges $\mathrm{q},-2 \mathrm{q}$ and $2 \mathrm{q}$ are placed on $x$-axis at a distance $x=0, x=\frac{3}{4} R$ and $x=R$ respectively from origin as shown. If $\mathrm{q}=2 \times 10^{-6} \mathrm{C}$ and $\mathrm{R}=2 \mathrm{~cm}$, the magnitude of net force experienced by the charge $-2 q$ is ___________ N.
Explanation:
A thin infinite sheet charge and an infinite line charge of respective charge densities $+\sigma$ and $+\lambda$ are placed parallel at $5 \mathrm{~m}$ distance from each other. Points 'P' and 'Q' are at $\frac{3}{\pi}$ m and $\frac{4}{\pi}$ m perpendicular distances from line charge towards sheet charge, respectively. '$\mathrm{E}_{\mathrm{P}}$' and '$\mathrm{E}_{\mathrm{Q}}$' are the magnitudes of resultant electric field intensities at point 'P' and 'Q', respectively. If $\frac{E_{p}}{E_{0}}=\frac{4}{a}$ for $2|\sigma|=|\lambda|$, then the value of $a$ is ___________.
Explanation:
$E_P = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{1}{4 \pi \varepsilon_0} \frac{2 \lambda}{3 / \pi}\right| = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{\lambda}{6 \varepsilon_0}\right| = \frac{\sigma}{6 \varepsilon_0}$
$E_Q = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{1}{4 \pi \varepsilon_0} \frac{2 \lambda}{4 / \pi}\right| = \left|\frac{\sigma}{2 \varepsilon_0} - \frac{\lambda}{8 \varepsilon_0}\right| = \frac{\sigma}{4 \varepsilon_0}$
Now we can find the ratio of $E_P$ to $E_Q$:
$\frac{E_P}{E_Q} = \frac{\frac{\sigma}{6 \varepsilon_0}}{\frac{\sigma}{4 \varepsilon_0}} = \frac{2}{3}$
As given in the question, $\frac{E_P}{E_0} = \frac{4}{a}$, and since $\frac{E_P}{E_Q} = \frac{2}{3}$, we can say $\frac{E_P}{E_Q} = \frac{E_P}{2E_Q} = \frac{4}{2 \times 3}$ = $\frac{4}{ 6}$ .
So, the value of $a$ is $\boxed{6}$.
64 identical drops each charged upto potential of $10 ~\mathrm{mV}$ are combined to form a bigger drop. The potential of the bigger drop will be __________ $\mathrm{mV}$.
Explanation:
The potential of each drop is given by:
$V = \frac{Kq}{r}$
where $K$ is the Coulomb constant, $q$ is the charge on each drop, and $r$ is the radius of each drop.
The radius of the bigger drop is:
$R = 4r$
since the 64 identical drops combine to form a bigger drop.
The total charge on the 64 identical drops is:
$Q = 64q$
The potential of the bigger drop is:
$V_{bigger} = \frac{KQ}{R} = \frac{K(64q)}{4r} = 16 \frac{Kq}{r} = 16V = 16 \times 10 \mathrm{~mV} = 160 \mathrm{~mV}$
Therefore, the potential of the bigger drop is $160 \mathrm{~mV}$.
As shown in the figure, a configuration of two equal point charges $\left(q_{0}=+2 \mu \mathrm{C}\right)$ is placed on an inclined plane. Mass of each point charge is $20 \mathrm{~g}$. Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height $\mathrm{h}=x \times 10^{-3} \mathrm{~m}$.
The value of $x$ is ____________.
(Take $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2}, g=10 \mathrm{~m} \mathrm{~s}^{-2}$ )
Explanation:
$ \begin{aligned} & \mathrm{mg} \sin \theta=\frac{1}{4 \pi \epsilon_0} \times \frac{\mathrm{q}_0^2}{\left(\mathrm{~h} \operatorname{cosec} 30^{\circ}\right)^2} \\\\ & \therefore \mathrm{h}^2=\frac{1}{4 \pi \epsilon_{\mathrm{o}}} \times \frac{\mathrm{q}_0^2}{\mathrm{mg} \operatorname{cosec} 30^{\circ}} \\\\ & =9 \times 10^9 \times \frac{\left(2 \times 10^{-6}\right)^2}{0.02 \times 10 \times 2} \\\\ & \therefore \mathrm{h}=3 \times 10^4 \times \frac{2 \times 10^{-6}}{0.2} \\\\ & =0.3 \mathrm{~m} \\\\ & =300 \mathrm{~mm} \end{aligned} $
An electron revolves around an infinite cylindrical wire having uniform linear charge density $2 \times 10^{-8} \mathrm{C} \mathrm{m}^{-1}$ in circular path under the influence of attractive electrostatic field as shown in the figure. The velocity of electron with which it is revolving is ___________ $\times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}$. Given mass of electron $=9 \times 10^{-31} \mathrm{~kg}$
Explanation:
$ \begin{aligned} & e E=\frac{\mathrm{mV}^2}{r} \\\\ & e \cdot \frac{2 \mathrm{~K} \lambda}{r}=\frac{\mathrm{mV}^2}{r} \\\\ & V=\sqrt{\frac{e \cdot 2 \mathrm{k} \lambda}{\mathrm{m}}} \\\\ & =\sqrt{\frac{1.6 \times 10^{-19} \times 2 \times 9 \times 10^9 \times 2 \times 10^{-8}}{9 \times 10^{-31}}} \\\\ & =8 \times 10^6 \mathrm{~m} / \mathrm{s} \end{aligned} $
Three concentric spherical metallic shells X, Y and Z of radius a, b and c respectively [a < b < c] have surface charge densities $\sigma,-\sigma$ and $\sigma$ respectively. The shells X and Z are at same potential. If the radii of X & Y are 2 cm and 3 cm, respectively. The radius of shell Z is _________ cm.
Explanation:
Given three concentric spherical shells X, Y, and Z with radii a, b, and c respectively, and with surface charge densities ( $\sigma$ ), ( $-\sigma$ ), and ( $\sigma$ ) respectively, we know that the potential at the surface of a sphere due to a uniform surface charge is given by:
$ V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} $
where ( $\epsilon_0$ ) is the permittivity of free space, ( Q ) is the total charge on the sphere, and ( r ) is the radius of the sphere.
However, in this case, the total charge on each sphere is given by its surface charge density ( $\sigma$ ) times its surface area ( $4\pi r^2$ ). Substituting this into the formula for ( Q ) gives:
$ Q = \sigma 4\pi r^2 $
So the potential at the surface of each sphere is given by:
$ V = \frac{1}{4\pi\epsilon_0} \frac{\sigma 4\pi r^2}{r} = \frac{\sigma r}{\epsilon_0} $
We are given that the potential at X and Z are the same. Thus:
$ V_X = V_Z $
Substituting the formula for the potential into this equation gives:
$ \frac{\sigma a}{\epsilon_0} = \frac{\sigma c}{\epsilon_0} $
This simplifies to:
$ a = c $
However, we also need to take into account the effect of the charge on shell Y on the potentials at X and Z. The potential at any point due to a charged shell is the same everywhere outside the shell, so we can add the potential due to shell Y at X to both sides of the equation. This gives:
$ \frac{\sigma a}{\epsilon_0} - \frac{\sigma b}{\epsilon_0} + \frac{\sigma c}{\epsilon_0} = \frac{\sigma a}{\epsilon_0} + \frac{\sigma c}{\epsilon_0} $
This simplifies to:
$ c(a - b + c) = a^2 - b^2 + c^2 $
Further simplification gives:
$ c(a - b) = a^2 - b^2 $
So:
$ c = a + b $
Given that the radii of X & Y are 2 cm and 3 cm, respectively, we have:
$ c = 2\, \text{cm} + 3\, \text{cm} = 5\, \text{cm} $
Therefore, the radius of shell Z is 5 cm.
An electric dipole of dipole moment is $6.0 \times 10^{-6} ~\mathrm{C m}$ placed in a uniform electric field of $1.5 \times 10^{3} ~\mathrm{NC}^{-1}$ in such a way that dipole moment is along electric field. The work done in rotating dipole by $180^{\circ}$ in this field will be ___________ $\mathrm{m J}$.
Explanation:
The work done $W$ in rotating an electric dipole in a uniform electric field is given by:
$W = pE(1 - \cos\theta)$,
where $p$ is the dipole moment, $E$ is the strength of the electric field, and $\theta$ is the angle the dipole is rotated through.
In this case, the dipole moment $p$ is $6.0 \times 10^{-6} ~\mathrm{C m}$, the electric field $E$ is $1.5 \times 10^{3} ~\mathrm{NC}^{-1}$, and the angle $\theta$ is $180^{\circ}$.
Substituting these values into the formula gives:
$W = 6.0 \times 10^{-6} ~\mathrm{C m} \times 1.5 \times 10^{3} ~\mathrm{NC}^{-1} \times (1 - \cos180^{\circ})$.
Since $\cos180^{\circ} = -1$, the equation becomes:
$W = 6.0 \times 10^{-6} ~\mathrm{C m} \times 1.5 \times 10^{3} ~\mathrm{NC}^{-1} \times (1 - (-1))$,
$W = 6.0 \times 10^{-6} ~\mathrm{C m} \times 1.5 \times 10^{3} ~\mathrm{NC}^{-1} \times 2$,
$W = 18.0 \times 10^{-3} ~\mathrm{J} = 18.0 ~\mathrm{mJ}$.
So the work done in rotating the dipole by $180^{\circ}$ in this field is 18.0 millijoules.
A cubical volume is bounded by the surfaces $\mathrm{x}=0, x=\mathrm{a}, y=0, y=\mathrm{a}, \mathrm{z}=0, z=\mathrm{a}$. The electric field in the region is given by $\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} x \hat{i}$. Where $\mathrm{E}_{0}=4 \times 10^{4} ~\mathrm{NC}^{-1} \mathrm{~m}^{-1}$. If $\mathrm{a}=2 \mathrm{~cm}$, the charge contained in the cubical volume is $\mathrm{Q} \times 10^{-14} \mathrm{C}$. The value of $\mathrm{Q}$ is ________________.
(Take $\epsilon_{0}=9 \times 10^{-12} ~\mathrm{C}^{2} / \mathrm{Nm}^{2}$)
Explanation:
$\begin{aligned} & \overrightarrow E = {E_0}x\widehat i \\\\ & \phi_{\mathrm{net}}=\phi_{\mathrm{ABCD}}=\mathrm{E}_0 \mathrm{a} \cdot \mathrm{a}^2 \\\\ & \frac{\mathrm{q}_{\mathrm{en}}}{\in_0}=\mathrm{E}_0 \mathrm{a}^3 \\\\ & \mathrm{q}_{\mathrm{en}}=\mathrm{E}_0 \in_0 \mathrm{a}^3 \\\\ & =4 \times 10^4 \times 9 \times 10^{-12} \times 8 \times 10^{-6} \\\\ & =288 \times 10^{-14} \mathrm{C} \\\\ & \therefore \mathrm{Q}=288\end{aligned}$
Two equal positive point charges are separated by a distance $2 a$. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge $\mathrm{q}_{0}$ becomes maximum is $\frac{a}{\sqrt{x}}$. The value of $x$ is __________.
Explanation:
$F_{P}=q_{0} E_{p}=q_{0} \frac{k q z}{\left(a^{2}+z^{2}\right)^{3 / 2}}$
$ \text { or } F_{P}=\frac{k q q_{0} z}{\left(a^{2}+z^{2}\right)^{3 / 2}} $
To maximize $\frac{d F_{P}}{d z}=0$
or $k q q_{0} \frac{\left(a^{2}+z^{2}\right)^{3 / 2}-z \frac{3}{2} \times 2 z\left(a^{2}+z^{2}\right)^{\frac{1}{2}}}{\left(a^{2}+z^{2}\right)^{3}}=0$
$\Rightarrow z=\frac{a}{\sqrt{2}}$
Expression for an electric field is given by $\overrightarrow{\mathrm{E}}=4000 x^{2} \hat{i} \frac{\mathrm{V}}{\mathrm{m}}$. The electric flux through the cube of side $20 \mathrm{~cm}$ when placed in electric field (as shown in the figure) is __________ $\mathrm{V} \mathrm{~cm}$.
Explanation:
$ \begin{aligned} & \text { So } \phi=E A \\\\ & =4000 \times(\cdot 2)^{2} \times \cdot 2 \times \cdot 2 \\\\ & =\frac{32}{5} \mathrm{Vm} \\\\ & =\frac{32}{5} \times 100 \mathrm{~V} \mathrm{~cm} \\\\ & =640 \mathrm{Vcm} \end{aligned} $
The value of $n$ is _________ (if dimension of cuboid is $1 \times 2 \times 3 \mathrm{~m}^{3}$ )
Explanation:
$ \begin{aligned} & =2(1)^{2} \times 2 \times 3 \\\\ & =12 \mathrm{Nm}^{2} / \mathrm{C} \end{aligned} $
Flux through planes parallel to $x-z=-4(2) \times$ Area
$ \begin{aligned} & =-4(2) \times 1 \times 3 \\\\ & =-24 \mathrm{Nm}^{2} / \mathrm{C} \end{aligned} $
Flux through planes parallel to $x-y=0$
$\Rightarrow \phi$ Total $=12-24=-12$
$\Rightarrow-12=\frac{q_{\mathrm{enc}}}{\varepsilon_{0}} \Rightarrow\left|q_{\mathrm{enc}}\right|=12 \varepsilon_{0}$
$\Rightarrow n=12$
For a charged spherical ball, electrostatic potential inside the ball varies with $r$ as $\mathrm{V}=2ar^2+b$.
Here, $a$ and $b$ are constant and r is the distance from the center. The volume charge density inside the ball is $-\lambda a\varepsilon$. The value of $\lambda$ is ____________.
$\varepsilon$ = permittivity of the medium
Explanation:
$V = 2a{r^2} + b$
$ \Rightarrow E = - {{dV} \over {dr}} = - 4ar$
$ \Rightarrow {1 \over {4\pi \varepsilon }}{Q \over {{r^2}}} = - 4ar$
$ \Rightarrow {Q \over {{4 \over 3}\pi {r^3}}} = 3 \times \varepsilon \times ( - 4a) = - 12a\varepsilon $
$ \Rightarrow \lambda = 12$
A point charge $q_1=4q_0$ is placed at origin. Another point charge $q_2=-q_0$ is placed at $x=12$ cm. Charge of proton is $q_0$. The proton is placed on $x$ axis so that the electrostatic force on the proton is zero. In this situation, the position of the proton from the origin is ___________ cm.
Explanation:
Let a proton having charge $q_0$ on the $x$ axis at distance $x$ from $q_2$ and $(12+x)$ distance from $q_1$.
Now, balance the force between them $\vec{F}_1+\vec{F}_2=0$
$ \frac{K 4 q_0\left(q_0\right)}{(12+x)^2}+\frac{K\left(-q_0\right)\left(q_0\right)}{x^2}=0 $
$ \Rightarrow $ $ \frac{4 K q_0^2}{(12+x)^2}=\frac{K\left(q_0\right)^2}{x^2} $
$ \Rightarrow $ $ \frac{4}{(12+x)^2}=\frac{1}{x^2} $
$ \Rightarrow $ $ \frac{2}{12+x}=\frac{1}{x} $
$ \Rightarrow $ $ 2 x=12+x $
$ \Rightarrow $ $ x=12 \mathrm{~cm} $
The charge $q_0$ is at distance of $24 \mathrm{~cm}$ from charge $q_1$ on $x$-axis.
$ \therefore $ Distance from origin is $12+12=24 \mathrm{~cm}$
A uniform electric field of 10 N/C is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The angle ($\theta$) of deviation of the path of electron as it comes out of the field is ___________ (in degree).
Explanation:
A stream of a positively charged particles having ${q \over m} = 2 \times {10^{11}}{C \over {kg}}$ and velocity ${\overrightarrow v _0} = 3 \times {10^7}\widehat i\,m/s$ is deflected by an electric field $1.8\widehat j$ kV/m. The electric field exists in a region of 10 cm along $x$ direction. Due to the electric field, the deflection of the charge particles in the $y$ direction is _________ mm.
Explanation:
$ \begin{aligned} & F_y=\frac{q E_y}{m} \\\\ & a_y=2 \times 10^{11} \times 1800 \\\\ & =36 \times 10^{13} \mathrm{~m} / \mathrm{s}^2 \\\\ & \text { Time }=\frac{10 \times 10^{-2}}{v_0}=\frac{0.1}{3 \times 10^7}=\left(\frac{1}{3} \times 10^{-8}\right) \mathrm{sec} \text {. } \\\\ & \therefore \quad y=\frac{1}{2} a t^2 \\\\ & \Rightarrow y=\frac{1}{2} \times 36 \times 10^{13} \times\left(\frac{1}{3} \times 10^{-8}\right)^2 \\\\ & =2 \times 10^{-3} \mathrm{~m} \\\\ & =2 \mathrm{~mm} \\\\ \end{aligned} $
Two electric dipoles of dipole moments $1.2 \times 10^{-30} \,\mathrm{Cm}$ and $2.4 \times 10^{-30} \,\mathrm{Cm}$ are placed in two different uniform electric fields of strengths $5 \times 10^{4} \,\mathrm{NC}^{-1}$ and $15 \times 10^{4} \,\mathrm{NC}^{-1}$ respectively. The ratio of maximum torque experienced by the electric dipoles will be $\frac{1}{x}$. The value of $x$ is __________.
Explanation:
${{{\rho _1}} \over {{\rho _2}}} = {{{\mu _1}{B_1}\sin 90} \over {{\mu _2}{B_2}\sin 90}}$
$ = {{1.2 \times {{10}^{ - 30}} \times 5 \times {{10}^4}} \over {2.4 \times {{10}^{ - 30}} \times 15 \times {{10}^4}}}$
$ = {1 \over 6}$
A long cylindrical volume contains a uniformly distributed charge of density $\rho \,\mathrm{Cm}^{-3}$. The electric field inside the cylindrical volume at a distance $x=\frac{2 \varepsilon_{0}}{\rho} \mathrm{m}$ from its axis is ________ $\mathrm{Vm}^{-1}$.
Explanation:
$E = {{\rho r} \over {2{\varepsilon _0}}}$
at $r = {{2{\varepsilon _0}} \over \rho }$
$E = {\rho \over {2{\varepsilon _0}}}\left( {{{2{\varepsilon _0}} \over \rho }} \right)$
$ = 1$
Three point charges of magnitude $5 \mu \mathrm{C}, 0.16 \mu \mathrm{C}$ and $0.3 \mu \mathrm{C}$ are located at the vertices $A, B, C$ of a right angled triangle whose sides are $A B=3 \mathrm{~cm}, B C=3 \sqrt{2} \mathrm{~cm}$ and $C A=3 \mathrm{~cm}$ and point $A$ is the right angle corner. Charge at point $\mathrm{A}$ experiences ____________ $\mathrm{N}$ of electrostatic force due to the other two charges.
Explanation:
${F_{AC}} = {{k(5 \times 0.3) \times {{10}^{ - 12}}} \over {9 \times {{10}^{ - 4}}}}$
${F_{AB}} = {{k(5 \times 0.16) \times {{10}^{ - 12}}} \over {9 \times {{10}^{ - 4}}}}$
${F_{net}} = {{k \times {{10}^{ - 12}}} \over {9 \times {{10}^{ - 4}}}}\sqrt {{{1.5}^2} + {{(0.8)}^2}} $
$ = {{{{10}^9} \times {{10}^{ - 12}}} \over {{{10}^{ - 4}}}} \times 1.7 = 17$
The volume charge density of a sphere of radius $6 \mathrm{~m}$ is $2 \,\mu \mathrm{C} \,\mathrm{cm}^{-3}$. The number of lines of force per unit surface area coming out from the surface of the sphere is _______________ $\times 10^{10} \,\mathrm{NC}^{-1}$.
[Given : Permittivity of vacuum $\epsilon_{0}=8.85 \times 10^{-12} \,\mathrm{C}^{2}\, \mathrm{~N}^{-1}-\mathrm{m}^{-2}$ )
Explanation:
$\rho$ = 2 $\mu$c/cm3
R = 6 m
Number of lines of force per unit area = Electric field at surface.
$ = {{KQ} \over {{R^2}}}$
$ = {1 \over {4\pi {\varepsilon _0}}}{{\rho {4 \over 3}\pi {R^3}} \over {{R^2}}}$
$ = {{\rho R} \over {3{ \in _0}}}$
$ = {{2 \times {{10}^{ - 6}} \times {{10}^6} \times 6} \over {3 \times 8.85 \times {{10}^{ - 12}}}}$
$ = 0.45197 \times {10^{12}}$
$ = 45.19 \times {10^{10}}$ N/C
$ \simeq 45 \times {10^{10}}$
Eight similar drops of mercury are maintained at 12 V each. All these spherical drops combine into a single big drop. The potential energy of bigger drop will be ____________ E. Where E is the potential energy of a single smaller drop.
Explanation:
$ \begin{aligned} &q_{i}=\mathrm{q}_{\mathrm{f}} \Rightarrow 8 \times\left(4 \pi \mathrm{E}_{0} \mathrm{R}\right) \times 12=\left(4 \pi E R^{1}\right) \times \mathrm{V}_{f} \\\\ &\Rightarrow 96 \mathrm{R}=\mathrm{V}_{\mathrm{f}} \mathrm{R}^{1} \\\\ &\text { And, } 8 \times \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi R^{1} 3 \\\\ &8=\left(\frac{R^{1}}{R}\right)^{3} \end{aligned} $
$ \begin{aligned} & \mathrm{R}^{1}=2 \mathrm{R} \end{aligned} $
From (i) & (ii), we get
So, $96 \mathrm{R}=\mathrm{V}_{\mathrm{f}} \times 2 \mathrm{R} \Rightarrow \mathrm{V}_{\mathrm{f}}=48$ Volt
$ \begin{aligned} &V_{f}=\frac{1}{2} C_{f} V_{f}^{2}=\frac{1}{2} \times\left(4 \pi \varepsilon_{0} \mathrm{R}^{1}\right) \mathrm{V}_{f}^{2} \\\\ &=\frac{1}{2} \times\left(4 \pi \varepsilon_{0} \times 2 \mathrm{R}\right) \times 48^{2} \\\\ &=\left(\frac{1}{2} \times 4 \pi \varepsilon_{0} R \times 12^{2}\right) \times \frac{48^{2} \times 2}{12^{2}}=32 \,E \end{aligned} $
27 identical drops are charged at 22V each. They combine to form a bigger drop. The potential of the bigger drop will be _____________ V.
Explanation:
Let the charge on one drop is q and its radius is r.
So for one drop $V = {{kq} \over r}$
For 27 drops merged new charge will be Q = 27 q and new radius is R = 3r
So new potential is
$V' = {{kQ} \over R} = 9{{kq} \over r} = 9 \times 22$ V
= 198 V
A charge $q$ is surrounded by a closed surface consisting of an inverted cone of height $h$ and base radius $R$, and a hemisphere of radius $R$ as shown in the figure. The electric flux through the conical surface is $\frac{n q}{6 \epsilon_{0}}$ (in SI units). The value of $n$ is _______.
Explanation:
Sol. From Gauss law,
$ \phi_{\text {hemisphere }}+\phi_{\text {Cone }}=\frac{\mathrm{q}}{\varepsilon_0} $ ............(1)
Total flux produced from $\mathrm{q}$ in $\alpha$ angle
$ \phi=\frac{\mathrm{q}}{2 \varepsilon_0}[1-\cos \alpha] $
For hemisphere, $\alpha=\frac{\pi}{2}$
$ \phi_{\text {hemisphere }}=\frac{\mathrm{q}}{2 \varepsilon_0} $
From equation (1)
$ \begin{aligned} & =\frac{\mathrm{q}}{2 \varepsilon_0}+\phi_{\text {cone }}=\frac{\mathrm{q}}{\varepsilon_0} \\\\ & \phi_{\text {cone }}=\frac{\mathrm{q}}{2 \varepsilon_0} \\\\ & \frac{4 \mathrm{q}}{6 \varepsilon_0}=\frac{\mathrm{q}}{2 \varepsilon_0} \\\\ & \mathrm{n}=3 \end{aligned} $
Explanation:
Net force on free charged particle
$F = {{k{q^2}} \over {{{(d + x)}^2}}} - {{k{q^2}} \over {{{(d - x)}^2}}}$
$F = - k{q^2}\left[ {{{4dx} \over {{{({d^2} - {x^2})}^2}}}} \right]$
$a = - {{4k{q^2}d} \over m}\left( {{x \over {{d^4}}}} \right)$
$a = - \left( {{{4k{q^2}} \over {m{d^3}}}} \right)x$
So, angular frequency
$\omega = \sqrt {{{4k{q^2}} \over {m{d^3}}}} $
$\omega = \sqrt {{{4 \times 9 \times {{10}^9} \times 10} \over {1 \times {{10}^{ - 6}} \times {1^3}}}} $
$\omega = 6 \times {10^8}$ rad/sec
$\omega = 6000 \times {10^5}$ rad/sec
D = e$-$x sin y $\widehat i$ $-$ e$-$x cos y $\widehat j$ + 2z $\widehat k$ C/m2
Explanation:
$Div.\,\overline E = {\rho \over {{\varepsilon _0}}}$
$ \Rightarrow div.\,\overline D = \rho $
$ \Rightarrow {\partial \over {\partial x}}\left( {{e^{ - x}}\sin y} \right) + {\partial \over {\partial y}}\left( { - {e^{ - x}}\cos y} \right) + {\partial \over {\partial z}}(2z) = \rho $
$\Rightarrow$ $\rho$ = 2 (a constant)
V = 2 $\times$ 10$-$9 m3
q = 2 $\times$ 2 $\times$ 10$-$9 = 4 nC
Explanation:
q = 8$\mu$C/g = 8 $\times$ 10$-$6 C/g = 8 $\times$ 10$-$3 C/kg
s = 10 cm = 0.1 m $\Rightarrow$ E = 100 V/m
We know that, acceleration, a = ${{force(F)} \over {mass(m)}}$
$\Rightarrow$ a = ${{qE} \over m}$ [$\because$ F = qE]
= ${{8 \times {{10}^{ - 6}} \times 100} \over {{{10}^{ - 3}}}}$ = 0.8 ms$-$2
As per question, when electric field is switched on, the body strikes to the wall and then returns back.
For one oscillation,
s = ut + ${1 \over 2}$ at2
$\Rightarrow$ 0.1 = ${1 \over 2}$ $\times$ 0.8 t2 [$\because$ u = 0]
$\Rightarrow$ 0.2 = 0.8 t2
$\Rightarrow$ ${2 \over 8}$ = t2 $\Rightarrow$ t2 = ${1 \over 4}$ $\Rightarrow$ t = ${1 \over 2}$
$\therefore$ Time period = 2 $\times$ ${1 \over 2}$ = 1 s
Therefore, if the collision of the body is perfectly elastic, the time period of motion will be 1s.
The total force on a 1C point charge, placed at the origin, is x $\times$ 103 N.
The value of x, to the nearest integer, is __________. [Take ${1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}$ Nm2/C2]
Explanation:
${F_{total}} = {{k{q_1}{q_2}} \over {r_1^2}} + {{k{q_1}{q_3}} \over {r_2^2}} + {{k{q_1}{q_4}} \over {r_3^2}} + .....$
$ = 9 \times {10^9} \times {10^{ - 6}}\left[ {1 + {{\left( {{1 \over 2}} \right)}^2} + {{\left( {{1 \over {{2^2}}}} \right)}^2} + {{\left( {{1 \over {{2^3}}}} \right)}^2} + {{\left( {{1 \over {{2^\infty }}}} \right)}^2}} \right]$
$ = 9 \times {10^9} \times {10^{ - 6}}\left[ {{1 \over {1 - {1 \over 4}}}} \right]$
$ \because $ $\left[ {{S_\infty } = {a \over {1 - r}}} \right]$ for G.P.
$ = 9 \times {10^3} \times {4 \over 3}$ = 12 $\times$ 103 N
Explanation:
$ = {{{E_0}} \over 5}\left( {2\widehat i + 3\widehat j} \right)\,.\,\left( {0.4\widehat i} \right)$
$ = {{4000} \over 5}\left( {2 \times 0.4} \right)$
$ = 640$ Nm2 C$-$1