JEE Physics Electric Charges and Fields MCQs Set 23

Answer: (d) \( \frac{\lambda}{2\pi\sqrt{\ell^2+r^2}} \)

Explanation:
1. Normal field from the semi-infinite thread: \( E_n=\int_\ell^\infty\frac{\lambda}{4\pi\epsilon_0}\frac{z\,dz}{(z^2+r^2)^{3/2}}=\frac{\lambda}{4\pi\epsilon_0\sqrt{\ell^2+r^2}} \).
2. With the image thread the normal field doubles, so \( |\sigma|=\epsilon_0\cdot2E_n=\frac{\lambda}{2\pi\sqrt{\ell^2+r^2}} \).

Teacher's Note:
1. At \( r=0 \) this reduces to the result at O.

Answer: (b) \( \frac{KQq}{3a^2} \) directed away from A, along AD

Explanation:
1. The forces from B and C act along BC in opposite directions and cancel (D is their midpoint).
2. The charge at A is at distance \( AD=\sqrt{(2a)^2-a^2}=\sqrt3a \), so its repulsion is \( \frac{KQq}{3a^2} \), directed from A to D.

Teacher's Note:
1. Use symmetry first: equal and opposite components cancel.

Answer: (a) the incentre of the triangle

Explanation:
1. Take a small element of a side that subtends angle \( \Delta\varphi \) at the incentre; its field equals that of an arc of the inscribed circle with the same linear density.
2. Summing over the sides is equivalent to a complete circular ring of charge, whose field at its centre is zero.

Teacher's Note:
1. The incentre is equidistant from all three sides, which makes the field cancel.

Answer: (d) \( -\frac{q\ell}{2\pi(\ell^2+r^2)^{3/2}} \)

Explanation:
1. Use the image charge \( -q \) at distance \( \ell \) on the other side. The normal field at the surface is twice the normal component of one charge: \( E_n=2\cdot\frac{q}{4\pi\epsilon_0(\ell^2+r^2)}\cdot\frac{\ell}{\sqrt{\ell^2+r^2}} \).
2. \( \sigma=-\epsilon_0E_n=-\frac{q\ell}{2\pi(\ell^2+r^2)^{3/2}} \).

Teacher's Note:
1. Surface charge density on a conductor equals \( \epsilon_0 \) times the normal field at its surface.

Answer: (a) \( \frac{Kq^2}{2b}-\frac{Kq^2}{2a} \)

Explanation:
1. Induced charges: \( -q \) on the inner surface and \( +q \) on the outer surface.
2. Potential at the centre \( =Kq\left(-\frac1a+\frac1b\right) \) from the induced charges (the charge's own potential is not counted in the work for bringing it).
3. Work \( =\frac12qV=\frac12 Kq^2\left(\frac1b-\frac1a\right) \), which is negative since \( a \lt b \).

Teacher's Note:
1. Work done equals the change in electrostatic energy of the whole system.

Answer: (a) \( \sigma_1=\sigma_4,\ \sigma_2=\epsilon_0E,\ \sigma_3=-\epsilon_0E \)

Explanation:
1. Inside a conductor the field is zero, and the field between the plates equals \( \frac{\sigma_2}{\epsilon_0} \) from the inner face of A.
2. Hence \( \sigma_2=\epsilon_0E \). Facing surfaces carry equal and opposite charge, so \( \sigma_3=-\sigma_2=-\epsilon_0E \).
3. The outer faces carry equal charge, \( \sigma_1=\sigma_4 \). The given \( E \) alone cannot fix their value.

Teacher's Note:
1. Facing surfaces of parallel plates carry equal and opposite charge; outer faces carry equal charge.

Answer: (d) stable, because the potential along this line is minimum at O

Explanation:
1. Along the line through O parallel to BC (coordinate \( x \) from O), \( V(x)\approx KQ\sqrt{\frac34}\left(3+\frac9{16}x^2\right) \).
2. \( V\propto x^2 \) has a minimum at \( x=0 \) with positive coefficient, so a like test charge is pushed back: stable for this direction.

Teacher's Note:
1. Equilibrium is stable along a line if the potential energy has a minimum there.

Answer: (c) \( N+1 \)

Explanation:
1. By symmetry the centre is a neutral point.
2. Each of the \( N \) axes of symmetry through a vertex carries one more neutral point, giving \( N \) further points.
3. Total \( =N+1 \).

Teacher's Note:
1. Use the symmetry axes of the polygon to locate points of zero field.

Answer: (d) execute simple harmonic motion about O

Explanation:
1. For small \( \delta \) the net force is \( F\approx\frac{9\sqrt3}{16}\frac{KQq\,\delta}{a^3} \), directed back towards A, i.e. towards O.
2. A restoring force proportional to the displacement gives simple harmonic motion.

Teacher's Note:
1. A force linear in displacement and opposite to it always gives SHM.

Answer: (b) \( \frac{R_1}{R_2} \) and \( 0 \)

Explanation:
1. Energy \( U=\frac{Q_1^2}{8\pi\epsilon_0R_1}+\frac{Q_2^2}{8\pi\epsilon_0R_2} \) with \( Q_1+Q_2 \) constant.
2. \( \frac{dU}{dQ_1}=0 \) gives \( \frac{Q_1}{R_1}=\frac{Q_2}{R_2} \), so \( \frac{Q_1}{Q_2}=\frac{R_1}{R_2} \).
3. Then the potentials \( \frac{Q_1}{4\pi\epsilon_0R_1} \) and \( \frac{Q_2}{4\pi\epsilon_0R_2} \) are equal, so the potential difference is \( 0 \).

Teacher's Note:
1. The energy is least when the two conductors are at the same potential; charge then flows no more.

Answer: (c) \( \frac{\rho_0r}{3\epsilon_0}\left(1-\frac{3r}{4R}\right) \)

Explanation:
1. Charge inside radius \( r \): \( \int_0^r4\pi r'^2\rho_0\left(1-\frac{r'}{R}\right)dr'=4\pi\rho_0\left(\frac{r^3}{3}-\frac{r^4}{4R}\right) \).
2. Gauss law: \( E=\frac{q_{enc}}{4\pi\epsilon_0r^2}=\frac{\rho_0r}{3\epsilon_0}\left(1-\frac{3r}{4R}\right) \).

Teacher's Note:
1. Use Gauss law with the enclosed charge found by integrating the density over thin shells.

Answer: (b) \( \frac{(2\sqrt2-1)q^2}{32\pi\epsilon_0\ell^2} \)

Explanation:
1. Images: \( -q \) at distance \( 2\ell \) along each plane normal, and \( +q \) at distance \( 2\sqrt2\,\ell \) diagonally.
2. Attractions \( \frac{Kq^2}{4\ell^2} \) along \( -x \) and \( -y \); repulsion \( \frac{Kq^2}{8\ell^2} \) along the diagonal.
3. Resultant along the diagonal: \( \sqrt2\left(\frac14-\frac1{8\sqrt2}\right)\frac{Kq^2}{\ell^2}=\frac{(2\sqrt2-1)Kq^2}{8\ell^2}=\frac{(2\sqrt2-1)q^2}{32\pi\epsilon_0\ell^2} \).

Teacher's Note:
1. Use the method of images with three image charges.

Answer: (c) \( 35 \) N to the right

Explanation:
1. Consider the system of shell plus point charge: its total charge is \( 1 \) C, so the external field gives a net force \( QE=20 \) N to the right on the whole system.
2. The point charge feels \( 15 \) N to the left, so the shell must feel \( 20-(-15)=35 \) N to the right.

Teacher's Note:
1. Apply Newton's third law and the total force on the combined system.

Answer: (d) \( \frac{\rho_0R}{9\epsilon_0} \) at \( r=\frac{2R}{3} \)

Explanation:
1. Inside, \( E=\frac{\rho_0}{3\epsilon_0}\left(r-\frac{3r^2}{4R}\right) \). \( \frac{dE}{dr}=0 \) gives \( 1-\frac{3r}{2R}=0 \), so \( r=\frac{2R}{3} \).
2. \( E_{max}=\frac{\rho_0}{3\epsilon_0}\left(\frac{2R}{3}-\frac{R}{3}\right)=\frac{\rho_0R}{9\epsilon_0} \).
3. Outside, \( E=\frac{\rho_0R^3}{12\epsilon_0r^2} \) decreases from \( \frac{\rho_0R}{12\epsilon_0} \), which is smaller.

Teacher's Note:
1. A maximum inside the ball is found by setting the derivative of the field to zero.

Answer: (c) \( -\frac{8Q}{11} \)

Explanation:
1. Let the final charges be \( q_A, q_B, q_C \) with \( q_A+q_C=Q \). Earthed B: \( V_B=0 \) gives \( 3q_A+3q_B+2q_C=0 \).
2. \( V_A=V_C \) gives \( \frac{q_A}{a}+\frac{q_B}{2a}+\frac{q_C}{3a}=\frac{q_A+q_B+q_C}{3a} \), so \( q_B=-4q_A \).
3. Then \( q_C=4.5q_A \), \( q_A=\frac{2Q}{11} \) and \( q_B=-\frac{8Q}{11} \).

Teacher's Note:
1. Use equal potentials for connected shells and zero potential for the earthed shell.

Answer: (a) \( 4\sqrt2\,q \)

Explanation:
1. The charged sphere polarises the neutral one: induced dipole \( p\propto E\propto\frac{q}{R^2} \).
2. The force on a dipole is \( p\frac{dE}{dR}\propto\frac{q}{R^2}\cdot\frac{q}{R^3}=\frac{q^2}{R^5} \).
3. For equal force, \( \frac{q'^2}{(2R)^5}=\frac{q^2}{R^5} \), so \( q'^2=32q^2 \) and \( q'=4\sqrt2\,q \).

Teacher's Note:
1. The force between a charge and an induced dipole falls as the fifth power of distance.

Answer: (b) a circle of radius \( 4 \) m whose centre lies \( 2 \) m from \( q \) on the side away from \( -2q \)

Explanation:
1. Put \( q \) at the origin and \( -2q \) at \( (6,0) \). Zero potential needs \( \frac{q}{r_1}=\frac{2q}{r_2} \), i.e. \( r_2=2r_1 \).
2. \( (x-6)^2+y^2=4(x^2+y^2) \) simplifies to \( (x+2)^2+y^2=16 \).
3. This is a circle of radius \( 4 \) m centred at \( (-2,0) \), which is \( 2 \) m from \( q \) on the side away from \( -2q \).

Teacher's Note:
1. The locus of points whose distances from two fixed points are in a fixed ratio (not \( 1 \)) is a circle (Apollonius circle).

Answer: (a) \( \frac{Qq}{6\epsilon_0d^2} \)

Explanation:
1. By Newton's third law the force equals the force of \( q \) on the plate. This is \( \frac{Q}{d^2}\times \) (flux of \( q \)'s field through the plate).
2. Surround \( q \) by a cube of edge \( d \): the plate is one face, so the flux through it is \( \frac{q}{6\epsilon_0} \).
3. \( F=\frac{Q}{d^2}\cdot\frac{q}{6\epsilon_0}=\frac{Qq}{6\epsilon_0d^2} \).

Teacher's Note:
1. The flux through one face of a cube enclosing a charge at its centre is one sixth of the total.

Answer: (b) \( 35 \) N to the left

Explanation:
1. The external field acts on the point charge with \( QE=20 \) N to the right (the shell screens only the field from outside).
2. Net force \( =-15 \) N, so the force from the shell is \( -15-20=-35 \) N, i.e. \( 35 \) N to the left.

Teacher's Note:
1. Net force is the sum of the force due to the external field and the force due to the induced charges.

Answer: (c) \( q_1\sin^2\frac{\alpha}{2}=q_2\sin^2\frac{\beta}{2} \)

Explanation:
1. Flux inside the cone of half-angle \( \alpha \) around \( q_1 \) is \( \frac{q_1}{2\epsilon_0}(1-\cos\alpha)=\frac{q_1}{\epsilon_0}\sin^2\frac\alpha2 \).
2. That flux all ends on \( -q_2 \) within the cone of half-angle \( \beta \), which carries \( \frac{q_2}{\epsilon_0}\sin^2\frac\beta2 \).
3. Equating gives \( q_1\sin^2\frac\alpha2=q_2\sin^2\frac\beta2 \).

Teacher's Note:
1. A bundle of field lines forms a tube of constant flux.

Answer: (a) \( \frac{\lambda}{2\pi\ell} \)

Explanation:
1. Field at O due to the semi-infinite thread: \( E=\int_\ell^\infty\frac{\lambda\,dz}{4\pi\epsilon_0z^2}=\frac{\lambda}{4\pi\epsilon_0\ell} \).
2. The image thread doubles the normal field at the surface: \( 2E \).
3. \( |\sigma|=\epsilon_0\cdot2E=\frac{\lambda}{2\pi\ell} \).

Teacher's Note:
1. A conducting plane doubles the normal field of the real charge because of the image charge.