Electric potential at a point '$\mathrm{P}$' due to a point charge of $5 \times 10^{-9} \mathrm{C}$ is $50 \mathrm{~V}$. The distance of '$\mathrm{P}$' from the point charge is:
(Assume, $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{+9} ~\mathrm{Nm}^{2} \mathrm{C}^{-2}$ )
Graphical variation of electric field due to a uniformly charged insulating solid sphere of radius $\mathrm{R}$, with distance $r$ from the centre O is represented by:
A dipole comprises of two charged particles of identical magnitude $q$ and opposite in nature. The mass 'm' of the positive charged particle is half of the mass of the negative charged particle. The two charges are separated by a distance '$l$'. If the dipole is placed in a uniform electric field '$\bar{E}$'; in such a way that dipole axis makes a very small angle with the electric field, '$\bar{E}$'. The angular frequency of the oscillations of the dipole when released is given by:
For a uniformly charged thin spherical shell, the electric potential (V) radially away from the centre (O) of shell can be graphically represented as -
Let $\sigma$ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different region $E_{I}, E_{I I}$ and $E_{I I I}$ are:
Which of the following correctly represents the variation of electric potential $(\mathrm{V})$ of a charged spherical conductor of radius $(\mathrm{R})$ with radial distance $(\mathrm{r})$ from the center?
$(\mathrm{I}: r < a, \mathrm{II}: a < r < b$, III: $r>b$ )
Electric field in a certain region is given by $\overrightarrow{\mathrm{E}}=\left(\frac{\mathrm{A}}{x^{2}} \hat{i}+\frac{\mathrm{B}}{y^{3}} \hat{j}\right) \text {. The } \mathrm{SI} \text { unit of } \mathrm{A} \text { and } \mathrm{B}$ are :
Two isolated metallic solid spheres of radii $\mathrm{R}$ and $2 \mathrm{R}$ are charged such that both have same charge density $\sigma$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $\sigma^{\prime}$. The ratio $\frac{\sigma^{\prime}}{\sigma}$ is :
A point charge $2\times10^{-2}~\mathrm{C}$ is moved from P to S in a uniform electric field of $30~\mathrm{NC^{-1}}$ directed along positive x-axis. If coordinates of P and S are (1, 2, 0) m and (0, 0, 0) m respectively, the work done by electric field will be
In a cuboid of dimension $2 \mathrm{~L} \times 2 \mathrm{~L} \times \mathrm{L}$, a charge $q$ is placed at the center of the surface '$\mathrm{S}$' having area of $4 \mathrm{~L}^{2}$. The flux through the opposite surface to '$\mathrm{S}$' is given by
A point charge of 10 $\mu$C is placed at the origin. At what location on the X-axis should a point charge of 40 $\mu$C be placed so that the net electric field is zero at $x=2$cm on the X-axis?
The electric potential at the centre of two concentric half rings of radii R$_1$ and R$_2$, having same linear charge density $\lambda$ is :
If two charges q$_1$ and q$_2$ are separated with distance 'd' and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic force?
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 identical metallic spheres $\mathrm{A}$ and $\mathrm{B}$ when placed at certain distance in air repel each other with a force of $\mathrm{F}$. Another identical uncharged sphere $\mathrm{C}$ is first placed in contact with $\mathrm{A}$ and then in contact with $\mathrm{B}$ and finally placed at midpoint between spheres A and B. The force experienced by sphere C will be:
A spherically symmetric charge distribution is considered with charge density varying as
$\rho(r)= \begin{cases}\rho_{0}\left(\frac{3}{4}-\frac{r}{R}\right) & \text { for } r \leq R \\ \text { zero } & \text { for } r>R\end{cases}$
Where, $r(r < R)$ is the distance from the centre O (as shown in figure). The electric field at point P will be:
Given below are two statements.
Statement I : Electric potential is constant within and at the surface of each conductor.
Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point.
In the light of the above statements, choose the most appropriate answer from the options given below.
A uniform electric field $\mathrm{E}=(8 \mathrm{~m} / \mathrm{e}) \,\mathrm{V} / \mathrm{m}$ is created between two parallel plates of length $1 \mathrm{~m}$ as shown in figure, (where $\mathrm{m}=$ mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of $2 \mathrm{~m} / \mathrm{s}$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be _________.
A charge of $4 \,\mu \mathrm{C}$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be :
Two identical positive charges $Q$ each are fixed at a distance of '2a' apart from each other. Another point charge $q_{0}$ with mass 'm' is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge $\mathrm{q}_{0}$ executes $\mathrm{SHM}$. The time period of oscillation of charge $\mathrm{q}_{0}$ will be :
Two uniformly charged spherical conductors $A$ and $B$ of radii $5 \mathrm{~mm}$ and $10 \mathrm{~mm}$ are separated by a distance of $2 \mathrm{~cm}$. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere $A$ and $B$ will be :
Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is :
If the electric potential at any point (x, y, z) m in space is given by V = 3x2 volt. The electric field at the point (1, 0, 3) m will be :
A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 $\times$ 105 NC$-$1. If the charge on the particle is 40 $\mu$C and the initial velocity is 200 ms$-$1, how much distance it will travel before coming to the rest momentarily :
Two point charges A and B of magnitude +8 $\times$ 10$-$6 C and $-$8 $\times$ 10$-$6 C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4 $\times$ 104 NC$-$1. The distance 'd' between the point charges A and B is :
Given below are two statements :
Statement I : A point charge is brought in an electric field. The value of electric field at a point near to the charge may increase if the charge is positive.
Statement II : An electric dipole is placed in a non-uniform electric field. The net electric force on the dipole will not be zero.
Choose the correct answer from the options given below :
The three charges q/2, q and q/2 are placed at the corners A, B and C of a square of side 'a' as shown in figure. The magnitude of electric field (E) at the corner D of the square, is :
If a charge q is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be :
Three identical charged balls each of charge 2 C are suspended from a common point P by silk threads of 2 m each (as shown in figure). They form an equilateral triangle of side 1m.
The ratio of net force on a charged ball to the force between any two charged balls will be :
Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 $\mu$C are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be :
Given below two statements : One is labelled as Assertion (A) and other is labelled as Reason (R).
Assertion (A) : Non-polar materials do not have any permanent dipole moment.
Reason (R) : When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it's individual atom or molecule coincides with the centre of the negative charge distribution.
In the light of above statements, choose the most appropriate answer from the options given below.
In the figure, a very large plane sheet of positive charge is shown. P1 and P2 are two points at distance l and 2l from the charge distribution. If $\sigma$ is the surface charge density, then the magnitude of electric fields E1 and E2 at P1 and P2 respectively are :
Two identical charged particles each having a mass 10 g and charge 2.0 $\times$ 10$-$7C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms$-$2]
A long cylindrical volume contains a uniformly distributed charge of density $\rho$. The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic energy of the particle is :