JEE Physics Electric Charges and Fields MCQs Set 18

Answer: (a) \( 8 \)

Explanation:
1. Only the faces perpendicular to the x-axis contribute. At \( x=0 \) the field is zero; at \( x=2 \) it is \( 2E_0 \).
2. Flux \( =2E_0\times(2\times2)=8E_0 \).
3. By Gauss law \( q=\epsilon_0\times8E_0 \), so \( n=8 \).

Teacher's Note:
1. Faces parallel to the field carry no flux. Apply Gauss law to the faces where the field is non-zero.

Answer: (d) \( 2 \)

Explanation:
1. In air \( \tan15^\circ=\frac{F}{mg} \). In the liquid the force is \( \frac FK \) and the effective weight is \( mg\left(1-\frac{0.8}{1.6}\right)=\frac{mg}{2} \).
2. The same angle means \( \frac{F}{K}=\frac F2 \), so \( K=2 \).

Teacher's Note:
1. The same angle means the same ratio of electric force to effective weight. Buoyancy halves the weight here.

Answer: (b) \( 16 \)

Explanation:
1. By Gauss law \( E(a)\cdot4\pi a^2=\frac{q}{\epsilon_0} \).
2. \( q=\epsilon_0(2\times2)(4\pi\times4)=64\pi\epsilon_0=4x\pi\epsilon_0 \), so \( x=16 \).

Teacher's Note:
1. The field has the same magnitude everywhere on the sphere, so flux \( =E\times \) area.

Answer: (b) (i) and (ii) only

Explanation:
1. The wire stays electrically neutral because electrons enter and leave at equal rates, so the enclosed net charge is zero before and after the switch is closed.
2. By Gauss law the flux is zero both times, so it remains unchanged (i) and it remains zero (ii).

Teacher's Note:
1. Flux depends only on the enclosed net charge. A steady current does not change it.

Answer: (d) \( 32 \)

Explanation:
1. The surface charge density is \( \sigma=\frac{Q}{4\pi R^2} \) and the outward electrostatic pressure on the surface is \( \frac{\sigma^2}{2\epsilon_0} \).
2. The net force on the smaller cap along the axis equals the pressure times its projected area \( \pi\left(R^2-h^2\right)=\frac{3\pi R^2}{4} \).
3. \( F=\frac{Q^2}{16\pi^2R^4}\cdot\frac{1}{2\epsilon_0}\cdot\frac{3\pi R^2}{4}=\frac{3Q^2}{128\pi\epsilon_0R^2}=\frac{3kQ^2}{32R^2} \), so \( p=32 \).

Teacher's Note:
1. The net force of a uniform pressure on a curved cap is pressure times the projected area.

Answer: (c) \( E \) rises linearly from zero at the centre to its maximum at \( r=R \) and then decreases as \( \frac{1}{r^2} \), continuously at \( r=R \)

Explanation:
1. Inside, \( E=\frac{\rho r}{3\epsilon_0}=\frac{Qr}{4\pi\epsilon_0R^3} \), which grows linearly with \( r \).
2. Outside, \( E=\frac{Q}{4\pi\epsilon_0r^2} \).
3. Both give \( \frac{Q}{4\pi\epsilon_0R^2} \) at \( r=R \), so the curve is continuous with a peak at \( r=R \).

Teacher's Note:
1. Contrast with a thin shell, where the field is zero inside and jumps at the surface. For a solid sphere the field is continuous.

Answer: (b) \( 3 \)

Explanation:
1. For small angles the separation satisfies \( d^3\propto q^2 \) (from \( \frac{d}{2L}=\frac{kq^2}{d^2mg} \)).
2. After one ball is discharged and the two are touched, each holds \( \frac q2 \), so \( b^3=\frac{a^3}{4} \).
3. \( b^3=\frac{108}{4}=27 \), so \( b=3 \) cm.

Teacher's Note:
1. In the small-angle limit the separation cubed is proportional to the square of the charge. Touching identical conductors shares the charge equally.

Answer: (a) \( E_2 \gt E_1 \gt E_3 \)

Explanation:
1. Sphere 1: \( P \) is outside, so \( E_1=\frac{kQ}{R^2} \).
2. Sphere 2: \( P \) is on the surface, so \( E_2=\frac{k(2Q)}{R^2} \).
3. Sphere 3: \( P \) is inside at \( R \) from the centre of a sphere of radius \( 2R \): \( E_3=\frac{k(4Q)R}{(2R)^3}=\frac{kQ}{2R^2} \).
4. Thus \( E_2 \gt E_1 \gt E_3 \).

Teacher's Note:
1. Outside and on the surface use \( \frac{kQ}{r^2} \). Inside a uniformly charged sphere \( E=\frac{kQr}{R^3} \).

Answer: (d) \( \frac{1}{4\pi\epsilon_0}\frac{2qQ}{a}\left(1-\frac{1}{\sqrt5}\right) \)

Explanation:
1. At the middle of side 1 the distances are \( a \) to each positive charge and \( \sqrt{a^2+4a^2}=\sqrt5a \) to each negative charge: \( V_i=k\left(\frac{2q}{a}-\frac{2q}{\sqrt5a}\right) \).
2. At the centre all four charges are at distance \( \sqrt2a \) and the charges sum to zero, so \( V_f=0 \).
3. \( K=Q(V_i-V_f)=\frac{1}{4\pi\epsilon_0}\frac{2qQ}{a}\left(1-\frac1{\sqrt5}\right) \).

Teacher's Note:
1. Kinetic energy gained equals \( Q \) times the fall in potential. At the centre the potential is zero by symmetry.

Answer: (d) \( 11 \)

Explanation:
1. \( \lambda=3\times10^{-8} \) C cm\( ^{-1}=3\times10^{-6} \) C m\( ^{-1} \). The force per unit length at separation \( r \) is \( \frac{\lambda^2}{2\pi\epsilon_0r} \).
2. \( W=\int_{1\,\text{cm}}^{2\,\text{cm}}\frac{\lambda^2}{2\pi\epsilon_0r}dr=\frac{\lambda^2}{2\pi\epsilon_0}\ln2=2(9\times10^9)(9\times10^{-12})(0.693)\approx0.112 \) J m\( ^{-1} \).
3. \( \frac{x}{100}=0.112 \) gives \( x\approx11 \).

Teacher's Note:
1. The field of a long line charge is \( \frac{\lambda}{2\pi\epsilon_0r} \), so the force on unit length of the other is \( \lambda E \). Integrate over the displacement.

Answer: (a) \( E_1\left(\frac{r_0}{2}\right)=2E_2\left(\frac{r_0}{2}\right) \)

Explanation:
1. Let the common value be \( E \). \( E_1\propto\frac1{r^2} \), \( E_2\propto\frac1r \) and \( E_3 \) is independent of \( r \).
2. At \( \frac{r_0}{2} \): \( E_1=4E \), \( E_2=2E \), \( E_3=E \). So \( E_1=2E_2 \) is true, while \( E_2=2E_3 \) (not \( 4E_3 \)) holds.
3. The equalities \( \frac{Q}{4\pi\epsilon_0r_0^2}=\frac{\sigma}{2\epsilon_0} \) and \( \frac{\lambda}{2\pi\epsilon_0r_0}=\frac{\sigma}{2\epsilon_0} \) give \( Q=2\pi\sigma r_0^2 \) and \( r_0=\frac{\lambda}{\pi\sigma} \), so the first two statements are false.

Teacher's Note:
1. Write the common value and scale each field to \( \frac{r_0}{2} \) using its distance dependence.

Answer: (c) \( 8 \)

Explanation:
1. A charge element \( dq \) at height \( z \) on the axis sends a flux \( \frac{dq}{4\pi\epsilon_0}\Omega \) through the disc, with solid angle \( \Omega=2\pi\left(1-\frac{z}{\sqrt{z^2+R^2}}\right) \).
2. \( \phi=\frac{\lambda}{2\epsilon_0}\int_0^\infty\left(1-\frac{z}{\sqrt{z^2+R^2}}\right)dz=\frac{\lambda}{2\epsilon_0}\left[z-\sqrt{z^2+R^2}\right]_0^\infty=\frac{\lambda R}{2\epsilon_0} \).
3. With \( \lambda=16\epsilon_0 \) and \( R=1 \): \( \phi=8 \) V m.

Teacher's Note:
1. Use the solid angle subtended by the disc at each point of the axis, then integrate along the thread.

Answer: (a) (ii) and (iii) only

Explanation:
1. Falling at constant speed: resistance \( R=mg=1.6\times10^{-18}\times10=1.6\times10^{-17} \) N (upward).
2. Rising at constant speed: resistance acts downward, so \( qE=mg+R=2mg=3.2\times10^{-17} \) N, directed upward. The drop is negative, so the field is downward (ii true).
3. \( E=\frac{3.2\times10^{-17}}{9.6\times10^{-19}}=33.3=\frac13\times10^2 \) N C\( ^{-1} \) (iii true).

Teacher's Note:
1. At constant speed the forces balance. The resistance reverses direction when the motion reverses. The force on a negative charge is opposite to the field.

Answer: (b) \( 9 \)

Explanation:
1. For a finite line charge of length \( \ell \) the field on the perpendicular bisector at distance \( b \) is \( E=\frac{kq}{b\sqrt{b^2+\ell^2/4}} \).
2. \( E=\frac{9\times10^9\times5\times10^{-9}}{2\sqrt{4+2.25}}=\frac{45}{2\times2.5}=9 \) V m\( ^{-1} \).

Teacher's Note:
1. Integrate the components of the field along the perpendicular direction only; the parallel components cancel by symmetry.

Answer: (a) \( 0 \)

Explanation:
1. The conducting shell screens the cavity: the field inside the cavity and in the metal is fixed only by the charges inside (here \( Q \) and the induced charge \( -Q \) on the inner surface), and these are spherically symmetric about \( Q \).
2. The outside charge \( q \) changes only the distribution of charge on the outer surface of the shell, which produces no field inside.
3. Hence the net force on \( Q \) is zero.

Teacher's Note:
1. A closed conducting shell shields its interior from outside charges, although the outside charge still feels a force from the shell's outer surface charge.

Answer: (c) \( \frac{Q}{2\pi a^2} \)

Explanation:
1. For \( a \lt r \lt b \): enclosed charge \( =Q+\int_a^r4\pi r'^2\frac{A}{r'}dr'=Q+2\pi A\left(r^2-a^2\right) \).
2. \( E=\frac{1}{4\pi\epsilon_0}\left[\frac{Q-2\pi Aa^2}{r^2}+2\pi A\right] \).
3. \( E \) is independent of \( r \) only if \( Q-2\pi Aa^2=0 \), so \( A=\frac{Q}{2\pi a^2} \).

Teacher's Note:
1. Write \( E \) as a sum of a \( \frac1{r^2} \) term and a constant; set the coefficient of the \( \frac1{r^2} \) term to zero.

Answer: (d) (i), (ii), (iii) and (iv)

Explanation:
1. The inner surface of \( B \) acquires \( -Q \) by induction (i true).
2. Earthing removes the \( +Q \) from the outer surface, so the outer surface carries no charge (iv true) and the field outside \( B \) is zero.
3. Inside \( A \) the field is zero (conductor). Between \( A \) and \( B \) the field is \( \frac{Q}{4\pi\epsilon_0r^2} \), which is not zero (iii true). So all four statements are correct.

Teacher's Note:
1. Earthing a shell removes the charge on its outer surface. Gauss law then shows no field outside.

Answer: (b) \( 60^\circ \)

Explanation:
1. Write \( \vec p=p(\cos\theta\,\hat i+\sin\theta\,\hat j) \). \( \vec T_1=\vec p\times\vec E_1=-pE\sin\theta\,\hat k \).
2. \( \vec T_2=\vec p\times\vec E_2=\sqrt3pE\cos\theta\,\hat k \).
3. \( \vec T_2=-\vec T_1 \) gives \( \sqrt3\cos\theta=\sin\theta \), so \( \tan\theta=\sqrt3 \) and \( \theta=60^\circ \).

Teacher's Note:
1. Use \( \vec T=\vec p\times\vec E \) with components. Equate the two torques with the given sign relation.

Answer: (b) \( -6a\epsilon_0 \)

Explanation:
1. \( E=-\frac{d\phi}{dr}=-2ar \).
2. Gauss law for a sphere of radius \( r \): \( E\cdot4\pi r^2=\frac{1}{\epsilon_0}\int_0^r4\pi r'^2\rho\,dr' \), so \( -8\pi ar^3\epsilon_0=\frac{4\pi\rho r^3}{3} \).
3. \( \rho=-6a\epsilon_0 \) (constant). Equivalently \( \rho=\epsilon_0\nabla\cdot\vec E=\epsilon_0\frac{1}{r^2}\frac{d}{dr}\left(-2ar^3\right) \).

Teacher's Note:
1. Use \( \vec E=-\nabla\phi \) and then Gauss law in differential form, \( \rho=\epsilon_0\nabla\cdot\vec E \).

Answer: (c) (ii) and (iv) only

Explanation:
1. The force on \( Q \) always points towards the origin, so \( Q \) oscillates about the origin along the x-axis and does not escape (iii false).
2. The potential energy is least at the origin, so the speed is greatest there (ii true).
3. The restoring force is \( \frac{2kqQx}{(x^2+a^2)^{3/2}} \), proportional to \( x \) only for \( x\ll a \). The amplitude is \( 2a \), which is not small, so the motion is oscillatory but not simple harmonic (i false, iv true).

Teacher's Note:
1. SHM needs a restoring force proportional to the displacement. Here that holds only for small displacements.

Answer: (a) \( 40 \) V cm\( ^{-1} \) along the \( +x \) direction

Explanation:
1. A uniform decrease means the potential is linear in \( x \), so the field is constant over the interval.
2. \( |E|=\frac{180-20}{4}=40 \) V cm\( ^{-1} \).
3. The field points in the direction of decreasing potential, that is \( +x \).

Teacher's Note:
1. A linear potential means a constant field equal to minus the slope. The field points towards falling potential.

Answer: (c) \( -5.31\times10^{-6} \) C

Explanation:
1. For \( r_A \lt r \lt r_B \) the enclosed charge is \( \epsilon_0\times(-4\times10^5)=-3.54\ \mu \)C.
2. This equals particle plus shell A: \( 1.77\ \mu\text{C}+q_A=-3.54\ \mu\text{C} \).
3. \( q_A=-5.31\ \mu \)C.

Teacher's Note:
1. The charge on a shell is the difference between the enclosed charges just outside and just inside it.