Electrostatics

A. Electrostatic field is zero inside a conductor.

B. Electric field at the surface of a charged conductor $=\frac{\sigma}{\varepsilon_0} \hat{n}$, depends on surface charge density ( $\sigma$ ).

C. The interior of a charged conductor cannot have any excess charge in the static situation.

D. At the surface of a charged conductor, the electrostatic field ⟂ surface.

E. The electrostatic potential is constant and can be non-zero everywhere inside a charged conductor.

Given:

Dipole moment, $ |\vec{P}| = 5 \times 10^{-6} \, \text{Cm} $

Electric field magnitude, $ |\vec{E}| = 4 \times 10^5 \, \text{N/C} $

Initial angle, $ \theta_i = 0^\circ $

Final angle, $ \theta_f = 60^\circ $

To calculate the change in potential energy $ \Delta U $ of the dipole:

$ \Delta U = U_f - U_i = -PE \cos \theta_f + PE \cos \theta_i $

Simplifying, we have:

$ \Delta U = PE \left( \cos \theta_i - \cos \theta_f \right) $

Substitute the known values:

$ \Delta U = 5 \times 10^{-6} \times 4 \times 10^5 \left(1 - \frac{1}{2}\right) $

Calculate further:

$ \Delta U = 5 \times 10^{-6} \times 4 \times 10^5 \times \frac{1}{2} $

$ \Delta U = 10 \times 10^{-6} \times 10^5 $

Which simplifies to:

$ \Delta U = 1 \, \text{J} $

Thus, the change in the potential energy of the dipole is $\boxed{1 \, \text{J}}$.

$\begin{aligned} F^{\prime} & =\frac{\frac{K q}{2} \frac{3 q}{4}}{r^2} \\ F^{\prime} & =\frac{3 F}{8} \end{aligned}$

To find the surface charge density of a charged metal cube, we need to determine the charge per unit area. The surface charge density, denoted by $\sigma$, is given by:

$\sigma = \frac{Q}{A}$

where:

• $ Q $ is the total charge on the cube
• $ A $ is the total surface area of the cube

The cube has a side length of $ 5 \mathrm{~cm} $, so each side of the cube is $ 5 \mathrm{~cm} $. Since 1 cm = 0.01 m, the side length in meters is:

$5 \mathrm{~cm} = 0.05 \mathrm{~m}$

The cube has 6 faces, and the area of one face is given by:

$\text{Area of one face} = \left( 0.05 \mathrm{~m} \right)^2 = 0.0025 \mathrm{~m}^2$

Therefore, the total surface area of the cube is:

$A = 6 \times 0.0025 \mathrm{~m}^2 = 0.015 \mathrm{~m}^2$

The total charge, $ Q $, is given as $ 6 \mu \mathrm{C} $. Converting this to Coulombs:

$6 \mu \mathrm{C} = 6 \times 10^{-6} \mathrm{~C}$

Now, plugging the values into the formula for surface charge density:

$\sigma = \frac{6 \times 10^{-6} \mathrm{~C}}{0.015 \mathrm{~m}^2} = 4 \times 10^{-4} \mathrm{~C} \mathrm{m}^{-2}$

Therefore, the surface charge density on the cube is:

$\sigma = 0.4 \times 10^{-3} \mathrm{~C} \mathrm{m}^{-2}$

So, the correct answer is:

Option D: $0.4 \times 10^{-3} \mathrm{C} \mathrm{m}^{-2}$

The electric potential (V) at a distance (r) from a point charge (q) is given by:

$V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r}$

where $\frac{1}{4 \pi \varepsilon_0}$ is Coulomb's constant.

In this case, we have:

• $q = 4 \times 10^{-7} \mathrm{C}$
• $r = 9 \mathrm{~cm} = 0.09 \mathrm{~m}$
• $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}$

Substituting these values into the equation for electric potential, we get:

$V = (9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}) \frac{(4 \times 10^{-7} \mathrm{C})}{(0.09 \mathrm{~m})}$

$V = 4 \times 10^4 \mathrm{~V}$

Therefore, the value of electric potential at a distance of $9 \mathrm{~cm}$ from the point charge $4 \times 10^{-7} \mathrm{C}$ is $4 \times 10^4 \mathrm{~V}$.

The correct answer is Option D.

For uniformly charged spherical shell,

$\begin{aligned} & V=\frac{k q}{R} \quad(\text { For } r \leq R) \\ & \therefore \quad V_C=V_P \\ & V_C-V_P=\text { Zero } \\ \end{aligned}$

The potential $V$ at any point, at distance $r$ from centre of dipole $=\frac{K P \cos \theta}{r^2}$

At axial point where $\theta=0^{\circ}, V=\frac{K P}{r^2}=\frac{9 \times 10^9 \times 4 \times 10^{-6}}{2^2}=9 \times 10^3 \mathrm{~V}$

At axial point where $\theta=180^{\circ}, V=\frac{-K P}{r^2}=-9 \times 10^3 \mathrm{~V}$

$\phi=\frac{\mathrm{q}_{\text {inside }}}{\varepsilon_0}$

only depends on charge enclosed by surface.

The Gaussian surface for the charge placed at the center of the cube will spread out equally through all sides of the cube. According to Gauss's Law, electric flux $\Phi$ through a closed surface is equal to the charge enclosed $Q$ divided by the permittivity of free space $\epsilon_0$:

$\Phi_{\text{total}} = \frac{Q}{\epsilon_0}$

Given that the cube has 6 faces, and due to symmetry, the flux through each face will be equal, the flux through any one face $\Phi_{\text{face}}$ is:

$\Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{Q}{6\epsilon_0}$

Since the charge is given in microcoulombs ($\mu\mathrm{C}$), we need to convert it to coulombs by multiplying with $10^{-6}$:

$\Phi_{\text{face}} = \frac{Q \times 10^{-6}}{6\epsilon_0}$

This matches option D, which means the correct flux through one face of the cube given a charge Q microcoulombs at the center is $\frac{Q}{6\epsilon_0} \times 10^{-6}$.

$ \begin{aligned} & \therefore V=\frac{K Q}{R} \\ & E=\frac{K Q}{r^2} \\ & E=\frac{V R}{r^2} \end{aligned} $

The torque τ experienced by an electric dipole in an electric field is given by the formula:

$\tau = pE\sin{\theta}$

where p is the electric dipole moment, E is the electric field intensity, and θ is the angle between the dipole and the electric field. The electric dipole moment p can be expressed as:

$p = qd$

where q is the charge on the dipole, and d is the dipole length.

We are given the following values:

• Torque τ = 4 N·m
• Electric field intensity E = $2 \times 10^5 ~\mathrm{NC}^{-1}$
• Angle θ = $30^{\circ}$
• Dipole length d = 2 cm = 0.02 m

We need to find the charge q on the dipole. Let's first solve for the electric dipole moment p:

$\tau = pE\sin{\theta}$

$ \Rightarrow $ $p = \frac{\tau}{E\sin{\theta}}$

Substituting the given values:

$p = \frac{4}{(2 \times 10^5) \sin{30^{\circ}}} = \frac{4}{(2 \times 10^5)(0.5)} = \frac{4}{10^5} = 4 \times 10^{-5} ~\mathrm{C~m}$

Now, let's solve for the charge q using the formula:

$ \Rightarrow $ $p = qd$

$q = \frac{p}{d}$

Substituting the values for p and d:

$q = \frac{4 \times 10^{-5}}{0.02} = 2 \times 10^{-3} ~\mathrm{C} = 2 ~\mathrm{mC}$

So, the magnitude of the charge on the dipole is 2 mC.

$\phi_{\mathrm{closed}}=0$

So, $\mathrm{\phi_{in}=\phi_{out}}$

Number of field lines entering is equal number of field lines leaving.

$\mathrm{v}=\frac{\mathrm{Kq}}{2 \times 10^{-2}}-\frac{\mathrm{Kq}}{8 \times 10^{-2}}$

$=\mathrm{Kq}\left[\frac{3}{8}\right] \times 10^{2}$

Work done $ = {q_0}\,.\,({V_0} - {V_\infty })$

$ = \left\{ {{{3kq} \over d} + \left( {{{ - 3kq} \over d}} \right)} \right\}{q_0}$

= Zero

dV = $-$$\overrightarrow E $ . d$\overrightarrow r $

dV = $-$Edr cos$\theta$

For equipotential surface,

dV = 0

cos$\theta$ = 0

$\Rightarrow$ $\theta$ = 90$^\circ$

Potential of conducting hollow sphere $ = {{KQ} \over R}$

Now, Q = same

$ \Rightarrow V \propto {1 \over R} \Rightarrow $ more the radius less will be the potential.

$\Rightarrow$ Hence potential would be more on smaller sphere.

For R >> L, arrangement is an electric dipole

$E = {{2p} \over {4\pi {\varepsilon _0}{R^3}}}$; where $p = qL$

$E \propto {1 \over {{R_3}}}$

As, $R$ be the radius of the ring. Consider a small strip of length $d l$ having charge $d q$ lying at an angle $\theta$.

$d l=R d \theta$

Charge on $d l=\lambda R d \theta$

Force at $1 \mathrm{~C}$ due to $\mathrm{dl}$

$=\frac{k \lambda R d \theta}{R^2}=\frac{k \lambda}{R} d \theta=d F$

We need to consider only the component $d F \cos \theta$, as the component $d F \sin \theta$ will cancel out because of the symmetrical element $d l$.

The total force on $1 \mathrm{C}$ is

$\begin{aligned} F & =\int_\limits{-\pi / 2}^{\pi / 2} d F \cos \theta \\ & =\frac{k \lambda}{R} \int_\limits{-\pi / 2}^{\pi / 2} \cos \theta d \theta \\ & =\frac{k \lambda}{R} \times 2=\frac{2 k \lambda}{R} \end{aligned}$

When the negative charge is shifted at a distance $x$ from the centre of the ring along its axis, then force acting on the point charge due to the ring.

$\begin{aligned} F & =q E ~(\text {towards centre}) \\\\ & =q \cdot \frac{k Q x}{\left(R^2+x^2\right)^{3 / 2}} \end{aligned}$

If $R >> x$, then $R^2+x^2 \simeq R^2$

and $\quad F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q q x}{R^3}$ (towards centre)

$\Rightarrow \quad a=\frac{F}{m}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q q x}{m R^3}$

Since, restoring force $F_E \propto x$, therefore motion of charge particle will be SHM.

Time period of SHM,

$T=\frac{2 \pi}{\omega}=\left[\frac{16 \pi^3 \varepsilon_0 R^3 m}{Q q}\right]^{1 / 2}\left[\because \omega^2=\frac{Q q}{4 \pi \varepsilon_0 R^3}\right]$

Negatively charged body contains more electron than its natural state of mass is slightly increased.

In Ist case, when charge +Q is situated at C

Electric potential energy of system

$U_1=\frac{1}{4 \pi \varepsilon_0} \frac{q(-q)}{2 L}+\frac{1}{4 \pi \varepsilon_0} \frac{(-q) Q}{L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}$

In Ind case, when charge $+Q$ is moved from $C$ to $D$

Electric potential energy of system in this case

$U_2=\frac{1}{4 \pi \varepsilon_0} \frac{q(-q)}{2 L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{3 L}+\frac{1}{4 \pi \varepsilon_0} \frac{(-q)(Q)}{L}$

Work done $(\Delta U)=U_2-U_1$

$\begin{aligned} = & \left(-\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2 L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{3 L}-\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}\right) \\ & -\left(-\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{2 L}-\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}+\frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}\right) \\ = & \frac{q Q}{4 \pi \varepsilon_0}\left[\frac{1}{3 L}-\frac{1}{L}\right]=\frac{q Q}{4 \pi \varepsilon_0} \frac{(1-3)}{3 L} \\ = & -\frac{2 q Q}{12 \pi \varepsilon_0 L}=-\frac{q Q}{6 \pi \varepsilon_0 L} \end{aligned}$

We know that, for an electric dipole

$\begin{aligned} E_{\text {axial }} & =\frac{1}{4 \pi \varepsilon_0}\left(\frac{2 p}{r^3}\right) \quad \text{..... (i)}\\ \text{and} \quad E_{\text {equatorial }} & =\frac{1}{4 \pi \varepsilon_0}\left(\frac{p}{r^3}\right) \quad \text{.... (ii)} \end{aligned}$

Hence, from Eqs. (i) and (ii), we get

$\frac{E_{\text {axil }}}{2}=E_{\text {equatorial }}$

Hence, $\quad E_{\text {equatorial }}=E / 2$

Reason is false as electric field due to dipole varies inversely as cube of distance, i.e.

$E \propto \frac{1}{r^3}$