JEE Physics Electric Charges and Fields MCQs Set 31

Answer: (a) \( \sqrt{\dfrac{2qE}{md}} \)

Explanation:
1. Restoring torque for a small angle: \( \tau=-qEd\,\theta \). Moment of inertia about the centre: \( I=2m\left(\dfrac d2\right)^2=\dfrac{md^2}{2} \).
2. \( \omega=\sqrt{\dfrac{qEd}{md^2/2}}=\sqrt{\dfrac{2qE}{md}} \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (b) A charge \( +q \) executes simple harmonic motion, while a charge \( -q \) continues to move in the direction of its displacement

Explanation:
1. At the midpoint the two fields cancel. Displace towards one line: the nearer line's field becomes stronger.
2. For \( +q \) the stronger field repels it back to the centre (restoring force, SHM). For \( -q \) the nearer line attracts it more strongly, so it keeps moving away from the midpoint.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (c) \( 1\times10^{-9}\ \text V \)

Explanation:
1. Just after release the proton has gravity (vertical) and the electric force (horizontal); \( 45^\circ \) means \( qE=mg \).
2. \( E=\dfrac{mg}{q}=\dfrac{1.67\times10^{-27}\times9.8}{1.6\times10^{-19}}=1.02\times10^{-7} \) V/m, so \( X=Ed=1.02\times10^{-7}\times10^{-2}\approx1\times10^{-9} \) V.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (d) \( \dfrac1{\varepsilon_0}\sigma^2R^2 \)

Explanation:
1. Electrostatic pressure on the shell: \( P=\dfrac{\sigma^2}{2\varepsilon_0} \).
2. The force pushing the hemispheres apart is pressure times the cross-section: \( F=\dfrac{\sigma^2}{2\varepsilon_0}\pi R^2\propto\dfrac{\sigma^2R^2}{\varepsilon_0} \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (a) \( 8.0\times10^{-19}\ \text C \)

Explanation:
1. Terminal velocity: \( 6\pi\eta rv=\dfrac43\pi r^3\rho g\Rightarrow r^2=\dfrac{9\eta v}{2\rho g}\), so \( r=4.29\times10^{-6} \) m (air buoyancy neglected).
2. Mass \( m=\dfrac43\pi r^3\rho=2.97\times10^{-13} \) kg. Balance: \( qE=mg\Rightarrow q=\dfrac{2.91\times10^{-12}}{3.635\times10^{6}}=8.0\times10^{-19} \) C.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (b) \( \dfrac{q^2}{54\pi\varepsilon_0R^2} \)

Explanation:
1. AB is a diameter, so \( \angle ACB=90^\circ \) and \( CB=AB\sin60^\circ=\sqrt3R \).
2. \( F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{(q/3)(2q/3)}{3R^2}=\dfrac{q^2}{54\pi\varepsilon_0R^2} \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (c) non-zero and uniform

Explanation:
1. Superpose a full sphere of density \( \rho \) and a small sphere of density \( -\rho \) in the cavity.
2. Inside: \( \vec E=\dfrac{\rho}{3\varepsilon_0}\vec r-\dfrac{\rho}{3\varepsilon_0}(\vec r-\vec d)=\dfrac{\rho\vec d}{3\varepsilon_0} \), constant and non-zero (d is the vector between the centres).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (d) the electric field at all points on the y-axis is along the +x direction

Explanation:
1. On the y-axis the field of each charge has components along y that cancel, and components along x that add, pointing from +q towards -q, i.e. along \( +x \).
2. The origin is at zero potential, so no work is needed to bring a test charge there. On the x-axis the field direction changes between regions. The dipole moment points from -q to +q, i.e. along \( -x \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (a) The balls bounce back to the bottom plate carrying the opposite charge to the one they went up with

Explanation:
1. On touching the top plate (at negative potential) a ball shares its potential and takes a negative charge.
2. The negative ball is now repelled by the top plate and attracted by the bottom one, so it returns carrying charge of the opposite sign to the one it went up with.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (b) proportional to \( V_0^2 \)

Explanation:
1. The charge a ball carries is proportional to the plate potential: \( q\propto V_0 \), and the field is \( E\propto V_0 \).
2. Time for one trip: \( t=\sqrt{\dfrac{2hm}{qE}}\propto\dfrac1{V_0} \). Charge transported per unit time \( \propto\dfrac{q}{t}\propto V_0\cdot V_0=V_0^2 \).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (c) P to 5; Q to 3; R to 1, 4; S to 2

Explanation:
1. Infinite plane: E independent of d (P to 5). Infinite line: \( E=\dfrac{\lambda}{2\pi\varepsilon_0d}\propto\dfrac1d \) (Q to 3).
2. Point charge: \( 1/d^2 \); a line dipole (two opposite parallel wires) behaves like a dipole in two dimensions, \( E\propto1/d^2 \): R to 1, 4. A point dipole: \( 1/d^3 \) (S to 2).

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (d) (iii) and (iv)

Explanation:
1. Inside the overlap, \( \vec E=\dfrac{\rho}{3\varepsilon_0}(\vec r_1-\vec r_2)=\dfrac{\rho}{3\varepsilon_0}\vec d \), where \( \vec d \) joins the centres.
2. This is uniform (same magnitude and direction) and non-zero, so the potential varies linearly: (iii) and (iv) are correct.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (a) \( -\dfrac{32}{25}\ \text{or}\ 4 \)

Explanation:
1. The centres are 3R apart. Point on the side towards the big sphere (R from its centre, inside it): small sphere \( \dfrac{\rho_1R}{12\varepsilon_0} \) and big sphere \( \dfrac{\rho_2R}{3\varepsilon_0} \) act in opposite directions, so \( \dfrac{\rho_1}{12}=\dfrac{\rho_2}{3}\Rightarrow\dfrac{\rho_1}{\rho_2}=4 \).
2. Point on the far side (5R from the big centre): both fields point the same way for equal signs, so \( \dfrac{\rho_1R}{12}=-\dfrac{8\rho_2R}{75}\Rightarrow\dfrac{\rho_1}{\rho_2}=-\dfrac{32}{25} \). Hence \( -\dfrac{32}{25} \) or 4.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (b) The angular momentum of the charge -q is constant

Explanation:
1. The Coulomb force is central, so its torque about +Q is zero and angular momentum is conserved.
2. In an ellipse the distance changes, so the speed, linear momentum and angular velocity all change.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (c) 3

Explanation:
1. The electric repulsion on each corner charge is of order \( \dfrac{q^2}{4\pi\varepsilon_0a^2} \); the film (two surfaces) pulls with a force of order \( \gamma a \).
2. Equilibrium: \( \dfrac{q^2}{a^2}\sim\gamma a\Rightarrow a^3\sim\dfrac{q^2}{\gamma} \), so \( a\propto(q^2/\gamma)^{1/3} \) and N = 3.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (d) be deflected in the direction of the field

Explanation:
1. X-rays eject photoelectrons from the metal, leaving the ball positively charged.
2. The force \( q\vec E \) on the positive charge is along \( \vec E \), so the ball is deflected in the direction of the field.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (a) different, the negatively charged sphere being slightly heavier

Explanation:
1. A positive charge Q is produced by removing electrons, which lowers the mass; a negative charge adds electrons and raises the mass.
2. So the two masses differ by \( 2\times\dfrac{Q}{e}m_e \) (a tiny difference), the negative sphere being heavier.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.

Answer: (b) 5.86 m/s

Explanation:
1. The Coulomb force from the fixed ball is always radial with constant magnitude \( F_e=\dfrac{9\times10^9\times10^{-12}}{0.8^2}=1.41\times10^{-2} \) N (outwards); it does no work as the ball moves on the circle.
2. At the top, tension zero: \( \dfrac{mv_t^2}{L}=mg-F_e\Rightarrow v_t^2=0.8\,(10-7.03)=2.38 \). Energy: \( v_b^2=v_t^2+4gL=2.38+32=34.4 \), so \( v_b=5.86 \) m/s.

Teacher's Note:
1. Write the governing law (Coulomb, Gauss, energy or RC charging) before substituting numbers.
2. Check the limiting case and the units of the final expression.