JEE Physics Electric Charges and Fields MCQs Set 16

Answer: (a) \( 4\ell\sin\frac\alpha2\sqrt{4\pi\varepsilon_0\,mg\sin\frac\alpha2} \)

Explanation:
1. With charge \( \frac q2 \) at each of A and B, \( AB=2\ell\sin\frac\alpha2 \) and the repulsion is \( F=\frac{(q/2)^2}{4\pi\varepsilon_0(2\ell\sin\frac\alpha2)^2} \).
2. Torque about O: weight acts at the middle of the rod, \( mg\frac\ell2\sin\alpha \); the electric force has lever arm \( \ell\cos\frac\alpha2 \) (the distance from O to line AB). Balance gives \( F=mg\sin\frac\alpha2 \).
3. Hence \( q^2=64\pi\varepsilon_0\ell^2mg\sin^3\frac\alpha2 \) and \( q=4\ell\sin\frac\alpha2\sqrt{4\pi\varepsilon_0mg\sin\frac\alpha2} \).

Teacher's Note:
1. Draw the charges and the field directions before writing equations; use symmetry to reduce the integrals.

Answer: (b) \( 2\pi\sqrt{\frac{2m\varepsilon_0}{3\lambda\sigma}} \)

Explanation:
1. The sheet gives a uniform field \( E=\frac\sigma{2\varepsilon_0} \). The rod is a dipole of moment \( p=\lambda\frac\ell2\cdot\frac\ell2=\frac{\lambda\ell^2}{4} \), pointing along the field, so it is in stable equilibrium.
2. Restoring torque for a small angle \( \theta \) is \( pE\theta=\frac{\lambda\sigma\ell^2}{8\varepsilon_0}\theta \); moment of inertia \( \frac{m\ell^2}{12} \).
3. So \( \omega^2=\frac{3\lambda\sigma}{2m\varepsilon_0} \) and \( T=2\pi\sqrt{\frac{2m\varepsilon_0}{3\lambda\sigma}} \).

Teacher's Note:
1. Draw the charges and the field directions before writing equations; use symmetry to reduce the integrals.

Answer: (c) \( \frac{16\sqrt3\,Kq_0^2}{a^2} \)

Explanation:
1. Originally each of the two forces is \( \frac{Kq_0^2}{a^2} \), at 60 degrees to each other, so the resultant is \( \sqrt3\frac{Kq_0^2}{a^2} \).
2. Doubling charges multiplies the force by 4; halving the side multiplies it by 4 again, giving \( 16\sqrt3\frac{Kq_0^2}{a^2} \), directed away from the other charges.

Teacher's Note:
1. Draw the charges and the field directions before writing equations; use symmetry to reduce the integrals.

Answer: (d) \( \frac{\sigma}{8\varepsilon_0} \)

Explanation:
1. Four identical quadrants make a full plane, which would give \( \frac\sigma{2\varepsilon_0} \) along z.
2. By symmetry each quadrant contributes the same z-component, so one quadrant gives \( \frac{\sigma}{8\varepsilon_0} \) (independent of z, as for the infinite plane).

Teacher's Note:
1. Draw the charges and the field directions before writing equations; use symmetry to reduce the integrals.

Answer: (a) \( 3.4\times10^{-8} \) C

Explanation:
1. Initially \( 2T_0\cos45^\circ=mg \). After the attraction F is added downward, \( 2(2T_0)\cos45^\circ=mg+F \), so \( F=mg=5.76\times10^{-3} \) N.
2. Coulomb's law: \( q=r\sqrt{\frac{F}{9\times10^9}}=0.042\times\sqrt{6.4\times10^{-13}}=3.4\times10^{-8} \) C.
3. (The new tension is \( \sqrt2\,mg=8.15\times10^{-3} \) N.)

Teacher's Note:
1. Draw the charges and the field directions before writing equations; use symmetry to reduce the integrals.

Answer: (b) \( \frac{q\lambda}{4\pi\varepsilon_0R} \)

Explanation:
1. The field of the ring on its axis is \( E(x)=\frac{q\,x}{4\pi\varepsilon_0(R^2+x^2)^{3/2}} \).
2. Force on the thread: \( F=\lambda\int_0^\infty E\,dx=\frac{\lambda q}{4\pi\varepsilon_0}\int_0^\infty\frac{x\,dx}{(R^2+x^2)^{3/2}}=\frac{\lambda q}{4\pi\varepsilon_0R} \).

Teacher's Note:
1. Draw the charges and the field directions before writing equations; use symmetry to reduce the integrals.

Answer: (c) \( E_{max}=\frac{\rho_0R}{9\varepsilon_0} \) at \( r=\frac{2R}3 \)

Explanation:
1. Gauss's law: \( E\,4\pi r^2=\frac{4\pi\rho_0}{\varepsilon_0}\int_0^r\left(r'^2-\frac{r'^3}{R}\right)dr' \), so \( E=\frac{\rho_0r}{3\varepsilon_0}\left(1-\frac{3r}{4R}\right) \) for \( r<R \).
2. Setting \( \frac{dE}{dr}=0 \) gives \( 1-\frac{3r}{2R}=0 \), so \( r_m=\frac{2R}{3} \).
3. Then \( E_{max}=\frac{\rho_0}{3\varepsilon_0}\cdot\frac{2R}3\cdot\frac12=\frac{\rho_0R}{9\varepsilon_0} \). (Outside, \( E=\frac{\rho_0R^3}{12\varepsilon_0r^2} \).)

Teacher's Note:
1. Draw the charges and the field directions before writing equations; use symmetry to reduce the integrals.

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

Explanation:
1. The charged sphere polarises the neutral one: induced dipole \( p\propto E\propto\frac{q}{R^2} \).
2. Force on a dipole in a non-uniform field: \( F\propto p\frac{dE}{dR}\propto\frac{q}{R^2}\cdot\frac{q}{R^3}=\frac{q^2}{R^5} \).
3. Doubling R divides F by \( 2^5=32 \), so \( q'^2=32q^2 \) and \( q'=4\sqrt2\,q \).

Teacher's Note:
1. Draw the charges and the field directions before writing equations; use symmetry to reduce the integrals.

Answer: (a) the incentre (centre of the inscribed circle)

Explanation:
1. A small piece of a side seen from the incentre at an angle \( \Delta\varphi \) carries charge \( \frac{\lambda r\Delta\varphi}{\cos^2\varphi} \) at distance \( \frac{r}{\cos\varphi} \), giving a field \( \frac{\lambda\Delta\varphi}{4\pi\varepsilon_0r} \) independent of the angle.
2. The same field would come from an arc of the inscribed circle of the same angle and density; the three sides together cover the whole circle symmetrically, so the net field is zero at the incentre.

Teacher's Note:
1. Draw the charges and the field directions before writing equations; use symmetry to reduce the integrals.

Answer: (b) \( \frac{Qq}{6\varepsilon_0d^2} \)

Explanation:
1. Place the charge at the centre of a cube of edge d: the plate is one face, and by Gauss's law the flux of q through it is \( \frac{q}{6\varepsilon_0} \).
2. The force on the plate is \( \sigma\Phi=\frac{Q}{d^2}\cdot\frac{q}{6\varepsilon_0} \) (the normal components add up as the flux), equal and opposite to that on q.
3. So \( F=\frac{Qq}{6\varepsilon_0d^2} \).

Teacher's Note:
1. Draw the charges and the field directions before writing equations; use symmetry to reduce the integrals.

Answer: (c) \( \frac{(2\sqrt2-1)q^2}{32\pi\varepsilon_0\ell^2} \)

Explanation:
1. Use images: \( -q \) at two points at distance \( 2\ell \) (attractive, perpendicular to each other) and \( +q \) at the opposite corner, at distance \( 2\sqrt2\ell \) (repulsive).
2. Two attractions: \( \sqrt2\cdot\frac{q^2}{4\pi\varepsilon_0(2\ell)^2} \) towards the corner. Repulsion: \( \frac{q^2}{4\pi\varepsilon_0\cdot8\ell^2} \) away from the corner.
3. Net \( \frac{q^2}{4\pi\varepsilon_0\ell^2}\left(\frac{\sqrt2}{4}-\frac18\right)=\frac{(2\sqrt2-1)q^2}{32\pi\varepsilon_0\ell^2} \).

Teacher's Note:
1. Draw the charges and the field directions before writing equations; use symmetry to reduce the integrals.

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

Explanation:
1. Use the image charge \( -q \) at distance \( \ell \) behind the plane; the normal field at the surface is \( E_n=2\cdot\frac{q}{4\pi\varepsilon_0(\ell^2+r^2)}\cdot\frac{\ell}{\sqrt{\ell^2+r^2}} \).
2. So \( |\sigma|=\varepsilon_0E_n=\frac{q\ell}{2\pi(\ell^2+r^2)^{3/2}} \). Integrating over the plane gives total induced charge \( -q \).

Teacher's Note:
1. Draw the charges and the field directions before writing equations; use symmetry to reduce the integrals.

Answer: (a) \( \left(\frac{9eaL}{2m}\right)^{1/3} \)

Explanation:
1. Acceleration \( =\frac{eE}{m}=\frac{eat}{mL} \), so \( v=\frac{eat^2}{2mL} \) and \( x=\frac{eat^3}{6mL} \).
2. Reaching the plate: \( L=\frac{eat^3}{6mL} \), i.e. \( t^3=\frac{6mL^2}{ea} \).
3. Then \( v^3=\frac{e^3a^3}{8m^3L^3}\cdot\frac{36m^2L^4}{e^2a^2}=\frac{9eaL}{2m} \).

Teacher's Note:
1. Draw the charges and the field directions before writing equations; use symmetry to reduce the integrals.