JEE Physics Electric Charges and Fields MCQs Set 15

Answer: (a) \( \frac{E^2q^2t^2}{2m} \)

Explanation:
1. Acceleration \( a=\frac{qE}{m} \), velocity \( v=\frac{qEt}{m} \).
2. \( K=\frac12mv^2=\frac{q^2E^2t^2}{2m} \).

Teacher's Note:
1. Use kinematics under constant force.
2. Kinetic energy grows as t squared.

Answer: (d) r/8

Explanation:
1. In a medium \( F'=\frac{kq_1q_2}{KR^2} \).
2. Set \( \frac{1}{16R^2}=\frac{4}{r^2} \), so \( R^2=\frac{r^2}{64} \) and \( R=\frac r8 \).

Teacher's Note:
1. Force in a dielectric is divided by K.
2. Equate the two expressions for the force.

Answer: (b) −q

Explanation:
1. Force on q from +4q: \( \frac{k\,4q\,q}{l^2} \) (repulsive, away from +4q).
2. Force from Q: \( \frac{kQq}{(l/2)^2}=\frac{4kQq}{l^2} \) must be opposite, so \( 4q+4Q=0 \) and \( Q=-q \).

Teacher's Note:
1. Q must attract q towards itself to cancel the repulsion.
2. Q lies at half the distance, which makes its force four times stronger per unit charge.

Answer: (d) increases

Explanation:
1. The upward electric force reduces the effective weight: \( g_{eff}=g-\frac{qE}{m}<g \).
2. \( T=2\pi\sqrt{l/g_{eff}} \) therefore increases.

Teacher's Note:
1. The electric force acts opposite to gravity here.
2. A smaller effective g gives a longer period.

Answer: (b) −2Q

Explanation:
1. The charge on the inner surface of a shell equals minus the total charge enclosed inside it: \( -(4Q-2Q)=-2Q \).

Teacher's Note:
1. Gauss's law inside the metal of the outer shell requires zero field.
2. Outer surface then has the remainder, -3Q.

Answer: (c) may be distributed uniformly on the surface

Explanation:
1. In electrostatic equilibrium all excess charge resides on the surface of a conductor.
2. It spreads uniformly only if no other charges or fields disturb the sphere; otherwise it is non-uniform, hence 'may be'.

Teacher's Note:
1. Charge cannot stay inside a conductor in equilibrium.
2. Nearby charges can make the surface distribution non-uniform.

Answer: (b) \( 32\times10^{-4} \) N m and \( 64\times10^{-4} \) J

Explanation:
1. Separation \( 2\times10^{-4}\ \text{m} \), so \( p=4\times10^{-8}\times2\times10^{-4}=8\times10^{-12}\ \text{C m} \).
2. \( \tau_{max}=pE=3.2\times10^{-3}\ \text{N m} \).
3. Work for 180°: \( W=2pE=6.4\times10^{-3}\ \text{J} \).

Teacher's Note:
1. Convert 2e-2 cm to metres: 2e-4 m.
2. Rotating from parallel to antiparallel changes U by 2pE.

Answer: (d) \( 6a\epsilon_0x \)

Explanation:
1. Field \( E=-\frac{d\varphi}{dx}=3ax^2 \).
2. Gauss's law: \( \rho=\epsilon_0\frac{dE}{dx}=6a\epsilon_0x \).

Teacher's Note:
1. Charge density is epsilon naught times the divergence of E.
2. In one dimension this is dE/dx.

Answer: (c) neither the electric field nor the electric potential is zero

Explanation:
1. On the axis \( E=\frac{2kp}{r^3} \) along the axis and \( V=\frac{kp}{r^2} \); both are non-zero at finite r.
2. Potential vanishes only on the equatorial line.

Teacher's Note:
1. The equatorial plane is the zero-potential plane.
2. Axial field is along the axis.

Answer: (c) \( -\phi \)

Explanation:
1. The closed surface (disc plus shell) encloses no charge, so the total flux is zero.
2. Field lines pass in through the curved surface and out through the disc, so the flux through the disc is \( +\phi \) and through the curved surface \( -\phi \).

Teacher's Note:
1. The charge is outside the closed surface.
2. Whatever lines enter must leave.

Answer: (d) \( \frac{\sqrt3\,q_0^2}{4\pi\epsilon_0a^2} \)

Explanation:
1. Each of the other two charges exerts \( F_0=\frac{q_0^2}{4\pi\epsilon_0a^2} \), and the two forces are 60° apart.
2. Resultant \( 2F_0\cos30^\circ=\sqrt3F_0 \).

Teacher's Note:
1. Add the two equal forces as vectors at 60 degrees.
2. The resultant bisects the angle between them.

Answer: (a) towards the plate B

Explanation:
1. Charges are induced on A: negative on the near side and positive on the far side.
2. The attractive force on the nearer negative charges is larger than the repulsive force on the farther positive charges, so A is attracted.

Teacher's Note:
1. A neutral conductor is always attracted by a charged body.
2. The nearer induced charge has the stronger interaction.

Answer: (c) 8 : 1

Explanation:
1. Initial force \( \propto 40\times20=800 \) (attractive).
2. After touching, each has \( \frac{40-20}{2}=10\ \mu\text{C} \), force \( \propto100 \) (repulsive).
3. Ratio \( \frac{800}{100}=8:1 \).

Teacher's Note:
1. Identical conductors share the total charge equally.
2. Compare the products of charges at equal distance.

Answer: (c) 7.6 m/s

Explanation:
1. The attraction towards the fixed charge acts along the thread towards the centre: \( F=\frac{9\times10^9\times(3\times10^{-6})^2}{1^2}=0.081\ \text{N} \); weight \( mg=0.1\ \text{N} \).
2. At the top, with tension just zero, both point towards the centre: \( \frac{mv_t^2}{r}=mg+F \), so \( v_t^2=\frac{0.181}{0.01}=18.1 \).
3. Energy conservation (the Coulomb energy does not change on the circle): \( v_b^2=v_t^2+4gr=18.1+40=58.1 \), so \( v_b\approx7.6\ \text{m/s} \).

Teacher's Note:
1. The Coulomb force is always radial, so it does no work on the ball.
2. The circle is completed if tension stays non-negative at the top.

Answer: (b) \( 2\pi\sqrt{\frac{l}{\sqrt{g^2+(qE/m)^2}}} \)

Explanation:
1. The weight mg (down) and the electric force qE (horizontal) are perpendicular, so the effective gravity is \( g_{eff}=\sqrt{g^2+(qE/m)^2} \).
2. \( T=2\pi\sqrt{l/g_{eff}} \).

Teacher's Note:
1. Add vertical and horizontal accelerations as vectors.
2. Use the effective g in the pendulum formula.

Answer: (b) \( \frac{\rho_0r^2}{4\epsilon_0} \)

Explanation:
1. Enclosed charge \( =\int_0^r\rho_0r'\,4\pi r'^2dr'=\pi\rho_0r^4 \).
2. \( E=\frac{\pi\rho_0r^4}{4\pi\epsilon_0r^2}=\frac{\rho_0r^2}{4\epsilon_0} \).

Teacher's Note:
1. Gauss's law with a spherical surface.
2. Outside the sphere the field is rho0 R^4/(4 eps0 r^2).

Answer: (a) may be neutral

Explanation:
1. Pair (2, 4) attracts, so 2 and 4 carry opposite charges.
2. Ball 1 attracts both 2 and 4. A charged ball cannot attract two oppositely charged bodies, so ball 1 must be neutral (attraction by induction).

Teacher's Note:
1. Only a neutral body is attracted by both a positive and a negative body.
2. It does not have to be a metal.

Answer: (b) \( \frac{q}{6\epsilon_0} \)

Explanation:
1. Each corner charge sends one eighth of its flux into the cube: \( \frac{q}{8\epsilon_0} \) per charge.
2. Total flux out of the cube \( =8\times\frac{q}{8\epsilon_0}=\frac q{\epsilon_0} \).
3. By symmetry each of the six faces gets \( \frac{q}{6\epsilon_0} \).

Teacher's Note:
1. A charge at a corner is shared by eight cubes.
2. Use symmetry to divide the total among the faces.

Answer: (a) \( \frac{q_1q_2}{56\pi\sqrt{14}\,\epsilon_0}\left(\hat j-2\hat i-3\hat k\right) \)

Explanation:
1. The vector from \( q_1 \) to \( q_2 \) is \( \vec r=(0-2)\hat i+(0+1)\hat j+(0-3)\hat k=-2\hat i+\hat j-3\hat k \) with \( |r|=\sqrt{14} \).
2. \( \vec F=\frac{q_1q_2}{4\pi\epsilon_0}\frac{\vec r}{r^3}=\frac{q_1q_2}{4\pi\epsilon_0\,14\sqrt{14}}(-2\hat i+\hat j-3\hat k) \), and \( 4\times14=56 \).

Teacher's Note:
1. Force on q2 points along the vector from q1 to q2 for like charges.
2. Divide by the cube of the distance because the vector is not a unit vector.

Answer: (d) \( 2.0\times10^{-3} \) N m

Explanation:
1. Dipole moment \( p=1.0\times10^{-6}\times2.0\times10^{-2}=2\times10^{-8}\ \text{C m} \).
2. \( \tau_{max}=pE=2\times10^{-8}\times10^5=2.0\times10^{-3}\ \text{N m} \).

Teacher's Note:
1. Maximum torque occurs when the dipole is perpendicular to the field.
2. Convert cm to m before multiplying.

Answer: (a) \( \sqrt{\frac{\sigma qd}{m\epsilon_0}} \)

Explanation:
1. Field near the plate \( E=\frac{\sigma}{2\epsilon_0} \) repels the particle (same sign). Work to be done against it: \( qEd=\frac{q\sigma d}{2\epsilon_0} \).
2. Need \( \frac12mv^2>\frac{q\sigma d}{2\epsilon_0} \), so \( v>\sqrt{\frac{\sigma qd}{m\epsilon_0}} \).

Teacher's Note:
1. An infinite non-conducting sheet gives a uniform field sigma over two epsilon naught.
2. Use the work-energy theorem.

Answer: (b) \( \frac{E_0a^3}{2} \)

Explanation:
1. Take area elements \( dA=a\,dx \) with the field along the area vector: \( \phi=\int_0^aE_0x\cdot a\,dx=E_0a\frac{a^2}{2}=\frac{E_0a^3}{2} \).

Teacher's Note:
1. The field varies with x, so integrate over strips parallel to the y-axis.
2. Each strip has area a dx.

Answer: (d) a net positive charge

Explanation:
1. Flux out through the top \( =+800A \); flux through the bottom (inward) \( =-400A \); side flux is zero (field parallel to surface).
2. Net outward flux \( =400A>0 \), so the enclosed charge is positive.

Teacher's Note:
1. Field lines entering count negative.
2. The unequal field strengths show that lines originate inside.

Answer: (a) \( \frac{\lambda_0}{4\epsilon_0R} \)

Explanation:
1. Component along the axis of symmetry: \( E=\int\frac{\lambda_0\cos\varphi\,R\,d\varphi}{4\pi\epsilon_0R^2}\cos\varphi=\frac{\lambda_0}{4\pi\epsilon_0R}\int_0^{2\pi}\cos^2\varphi\,d\varphi=\frac{\lambda_0}{4\epsilon_0R} \) (the perpendicular component integrates to zero).

Teacher's Note:
1. Resolve the field of each element along and across the direction phi = 0.
2. The integral of cos squared over a full turn is pi.

Answer: (b) \( \frac{5q}{6\epsilon_0} \)

Explanation:
1. The charge is at the centre, so each of the six faces would carry \( \frac{q}{6\epsilon_0} \).
2. With one face open, five faces remain: \( \frac{5q}{6\epsilon_0} \).

Teacher's Note:
1. Symmetry gives equal flux per face.
2. The open face lets one sixth escape.