JEE Physics Electric Charges and Fields MCQs Set 21

Answer: (a) \( \sqrt{\dfrac{3qE_0a}{m}} \)

Explanation:
1. The constant force \( qE_0 \) acts downward, like an effective gravity.
2. The top of the circle is the critical point: \( \dfrac{mv_t^2}{a}\ge qE_0 \) so \( v_t^2\ge\dfrac{qE_0a}{m} \).
3. From (a,0) to the top (0,a) the sphere rises by a against the force, so \( v^2=v_t^2+\dfrac{2qE_0a}{m} \).
4. Minimum \( v=\sqrt{3qE_0a/m} \).

Teacher's Note:
1. Treat the uniform field as an effective gravity and use the standard vertical circle condition.
2. The start point is level with the centre, not at the bottom, so the result differs from the familiar root-5 case.

Answer: (b) a charged particle may move along an electric line of force

Explanation:
1. The acceleration is along the line of force at each point, but the velocity need not be.
2. If released from rest on a straight line of force, or given an initial velocity along it, the particle moves along it.
3. Hence it may, but need not, move along a line of force (curved lines are generally not followed).

Teacher's Note:
1. Field lines give the direction of the force, not the path.
2. A path coincides with a field line only in special cases.

Answer: (b) E is conserved but p is not

Explanation:
1. The fixed support exerts an external force on Y, so the momentum of X plus Y is not conserved.
2. The electrostatic force is conservative and the fixed Y does no work, so the total mechanical energy is conserved.

Teacher's Note:
1. External impulses from the support spoil momentum conservation.
2. Static supports do no work, so energy is unaffected.

Answer: (c) \( \sqrt{\dfrac{Qqm}{4\pi\epsilon_0 r}} \)

Explanation:
1. Energy: \( \tfrac12mv^2=\dfrac{kQq}{r}-\dfrac{kQq}{2r}=\dfrac{Qq}{8\pi\epsilon_0r} \) so \( v^2=\dfrac{Qq}{4\pi\epsilon_0mr} \).
2. The agent holds Q fixed, so its impulse equals the momentum gained by q: \( mv=\sqrt{\dfrac{Qqm}{4\pi\epsilon_0r}} \).

Teacher's Note:
1. The agent must cancel the force from q on Q at every instant, so its impulse equals the momentum given to q.
2. Use energy conservation to get the speed.

Answer: (a) \( \dfrac{q}{2r}\sqrt{\dfrac{1}{\pi\epsilon_0r}} \)

Explanation:
1. At maximum extension the separation is 2r and the particles are momentarily at rest.
2. Energy: \( \dfrac{q^2}{4\pi\epsilon_0r}=\dfrac{q^2}{4\pi\epsilon_0(2r)}+\tfrac12kr^2 \).
3. So \( \tfrac12kr^2=\dfrac{q^2}{8\pi\epsilon_0r} \), \( k=\dfrac{q^2}{4\pi\epsilon_0r^3}=\dfrac{q}{2r}\sqrt{\dfrac1{\pi\epsilon_0r}} \).

Teacher's Note:
1. The electric potential energy released by the extra separation is stored in the spring.
2. Simplify the square root carefully: the constant 4 comes out as a factor 2.

Answer: (d) \( \text{may be equal to } V/x_0 \)

Explanation:
1. The x-component of the field is \( E_x=-\dfrac{\Delta V}{\Delta x}=\dfrac{2V}{2x_0}=\dfrac{V}{x_0} \).
2. The field may also have y and z components, so the magnitude is at least \( V/x_0 \).
3. It may therefore equal \( V/x_0 \) but cannot be less.

Teacher's Note:
1. The potential is given only along a line, so only one component of the field is fixed.
2. The magnitude can never be less than any single component.

Answer: (c) E can be zero at a point where V is not zero

Explanation:
1. E is the negative gradient of V, so E depends on the slope of V, not on its value.
2. Example: inside a uniformly charged conducting shell, E = 0 but V is non-zero.
3. Similarly, at the mid-point of an electric dipole's equatorial line V = 0 but E is not.

Teacher's Note:
1. Zero field means constant potential, not zero potential.
2. The converse also fails: V = 0 does not imply E = 0.

Answer: (a) the x-component of the field at C is along the positive x-axis

Explanation:
1. \( E_x=-dV/dx \).
2. At A the slope is positive, so \( E_x<0 \); at B the slope is zero, so \( E_x=0 \); at C the slope is negative, so \( E_x>0 \).

Teacher's Note:
1. The field points in the direction of decreasing potential.
2. At a maximum of V the field component vanishes.

Answer: (c) \( \dfrac{2\sqrt2\,qR}{\pi} \)

Explanation:
1. The dipole moment of a neutral system is \( q\,\vec r_{cm} \), where the centroid of the arc is at \( \dfrac{R\sin(\alpha/2)}{\alpha/2} \) from the centre.
2. With \( \alpha=\pi/2 \): \( r_{cm}=\dfrac{R\sin45^\circ}{\pi/4}=\dfrac{2\sqrt2R}{\pi} \).
3. So \( p=\dfrac{2\sqrt2qR}{\pi} \).

Teacher's Note:
1. Only the position of the arc's centroid matters because the other charge is at the origin.
2. Use the arc-centroid formula R sin(half-angle) over half-angle.

Answer: (d) \( \dfrac{\sqrt7}{32\pi\epsilon_0}\ \text{N/C} \)

Explanation:
1. With \( \hat r=\hat i \) and \( r=2 \): \( \vec E=\dfrac{k}{r^3}[3(\vec p\cdot\hat r)\hat r-\vec p]=\dfrac{k}{8}[3\hat i-\hat i-\sqrt3\hat j]=\dfrac{k}{8}(2\hat i-\sqrt3\hat j) \).
2. Magnitude \( \dfrac{k}{8}\sqrt{4+3}=\dfrac{\sqrt7}{8}\cdot\dfrac{1}{4\pi\epsilon_0}=\dfrac{\sqrt7}{32\pi\epsilon_0} \).

Teacher's Note:
1. Use the general dipole field formula, not just the axial or equatorial special cases.
2. Resolve p along and perpendicular to the position vector.

Answer: (a) \( \dfrac{6kp_1p_2}{x^4} \), attractive

Explanation:
1. The axial field of p₁ at p₂ is \( E=\dfrac{2kp_1}{x^3} \).
2. The force on p₂ is \( p_2\left|\dfrac{dE}{dx}\right|=\dfrac{6kp_1p_2}{x^4} \).
3. With both moments pointing the same way, the nearer ends are opposite charges, so the force is attractive.

Teacher's Note:
1. Force on a dipole is p times the gradient of the field.
2. Head-to-tail aligned dipoles attract.

Answer: (c) unstable along the line joining the charges and stable perpendicular to that line

Explanation:
1. Along the line, a displacement brings the negative charge closer to one positive charge, whose pull then exceeds the other, so the charge is dragged further away: unstable.
2. Perpendicular to the line, both positive charges pull the negative charge back towards the line, so it returns: stable.

Teacher's Note:
1. Earnshaw's theorem forbids stability in all directions for electrostatic forces alone.
2. Test each direction separately.

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

Explanation:
1. The pair with the largest product (4Q and 16Q) must be farthest apart, so they take the ends.
2. With Q at x from 4Q: \( U\propto\dfrac4x+\dfrac{16}{9-x} \); minimising gives \( 9-x=2x \), x = 3 cm from 4Q, i.e. 6 cm from 16Q. So (ii) is correct.
3. At that point the fields cancel: \( \dfrac{4}{3^2}=\dfrac{16}{6^2} \) so the field is zero, (iii) is correct.

Teacher's Note:
1. Put the strongly repelling pair as far apart as possible.
2. Minimum energy implies zero force on the middle charge.

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

Explanation:
1. At closest approach the separation is momentarily constant, so both particles have the same velocity; masses may differ, so momenta need not be equal.
2. Total momentum is conserved, so the momentum gained by B equals that lost by A.
3. Part of A's lost kinetic energy is stored as electrostatic potential energy, so B has gained less than A has lost.

Teacher's Note:
1. Closest approach means zero relative velocity.
2. Energy is shared between kinetic and potential forms.

Answer: (a) (i) and (iv) only

Explanation:
1. Charge is a property of matter, whereas matter can be neutral, so (i) is correct.
2. Charge is invariant with speed, so (ii) is wrong.
3. Charge sharing depends on the relative sizes of the bodies, so (iii) is wrong in general.
4. Attraction can also occur between a charged and a neutral body, so only repulsion proves electrification, making (iv) correct.

Teacher's Note:
1. Charge is relativistically invariant.
2. Attraction alone is ambiguous because induction also gives attraction.

Answer: (b) 2

Explanation:
1. In air: \( \tan\theta=\dfrac{F}{mg} \). In liquid: \( \tan\theta=\dfrac{F/K}{mg(1-\sigma/\rho)} \).
2. Same angle gives \( K=\dfrac{1}{1-\sigma/\rho}=\dfrac{1}{1-0.8/1.6}=2 \).

Teacher's Note:
1. Both the electric force and the effective weight change in the liquid.
2. Buoyancy reduces the apparent weight by the factor (1 − σ/ρ).

Answer: (c) \( \dfrac{16kQq}{3\sqrt3\,d^2} \)

Explanation:
1. At distance y from the mid-point, \( F=\dfrac{2kQq\,y}{(y^2+d^2/4)^{3/2}} \).
2. Setting \( dF/dy=0 \) gives \( y=\dfrac{d}{2\sqrt2} \).
3. Substituting: \( F_{max}=\dfrac{16kQq}{3\sqrt3\,d^2} \).

Teacher's Note:
1. The force is zero at the mid-point and at infinity, so a maximum lies in between.
2. Differentiate and set to zero.

Answer: (b) \( \dfrac{qQ}{8\pi^2\epsilon_0r^2} \)

Explanation:
1. Take a small element subtending angle \( d\theta \): charge \( dq=\dfrac{Q}{2\pi}d\theta \), force from q \( dF=\dfrac{q\,dq}{4\pi\epsilon_0r^2} \) radially outward (for like charges).
2. Balance: \( 2T\sin\dfrac{d\theta}{2}=dF \), so \( T\,d\theta=\dfrac{qQ\,d\theta}{8\pi^2\epsilon_0r^2} \).
3. Hence \( T=\dfrac{qQ}{8\pi^2\epsilon_0r^2} \).

Teacher's Note:
1. Ignore the ring's own self-repulsion, which is not asked.
2. Balance the outward electric force with the net inward tension on a small element.

Answer: (d) \( 1.8\times10^{5}\ \text{s} \)

Explanation:
1. Initially \( U_i=3\dfrac{kq^2}{1} \). Finally the moved charge is 0.5 m from each of the others, which are 1 m apart: \( U_f=kq^2\left[\dfrac{2}{0.5}+\dfrac11\right]=5kq^2 \).
2. \( \Delta U=2kq^2=2\times9\times10^9\times0.01=1.8\times10^8\ \text{J} \).
3. \( t=\dfrac{1.8\times10^8}{1000}=1.8\times10^5\ \text{s} \).

Teacher's Note:
1. Only the energy difference counts.
2. Power is energy per unit time.

Answer: (b) 1.2 J

Explanation:
1. The distances from the origin are 0.03 m and 0.04 m. \( V_A=\dfrac{9\times10^9\times8\times10^{-3}}{0.03}=2.4\times10^6\ \text{V} \), \( V_B=1.8\times10^6\ \text{V} \).
2. \( W=q(V_B-V_A)=(-2\times10^{-9})(1.8-2.4)\times10^6=+1.2\ \text{J} \).

Teacher's Note:
1. Electrostatic work depends only on end points.
2. Mind the sign of the test charge.

Answer: (a) \( \dfrac{4a}{3} \)

Explanation:
1. Axial potential: \( V(z)=\dfrac{\sigma}{2\epsilon_0}\left(\sqrt{a^2+z^2}-z\right) \), and \( \dfrac{q\sigma}{2\epsilon_0}=2mg \).
2. Energy balance: \( mgH+q[V(H)-V(0)]=0 \) gives \( H+2\left[\sqrt{a^2+H^2}-H-a\right]=0 \).
3. Hence \( 2\sqrt{a^2+H^2}=2a+H \), \( 3H^2=4aH \) and \( H=4a/3 \).

Teacher's Note:
1. The disc repels the particle, so the electric energy rises as it approaches.
2. Square the equation once and discard the zero root.

Answer: (b) \( \dfrac{a}{\sqrt3} \)

Explanation:
1. Axial field of the disc: \( E=\dfrac{\sigma}{2\epsilon_0}\left(1-\dfrac{z}{\sqrt{a^2+z^2}}\right) \).
2. Equilibrium: \( qE=mg \) gives \( 2\left(1-\dfrac{z}{\sqrt{a^2+z^2}}\right)=1 \).
3. So \( \dfrac{z}{\sqrt{a^2+z^2}}=\dfrac12 \) and \( z=\dfrac{a}{\sqrt3} \).

Teacher's Note:
1. Balance the electric force with the weight.
2. Solve z over root(a squared plus z squared) equals one half.

Answer: (d) \( \dfrac{Q}{\sqrt{2\pi\epsilon_0mR}} \)

Explanation:
1. Potential at the centre: \( V=\dfrac{Q}{4\pi\epsilon_0R} \); far away V = 0.
2. Gain in kinetic energy \( \tfrac12mv^2=|{-Q}|V=\dfrac{Q^2}{4\pi\epsilon_0R} \).
3. So \( v=\sqrt{\dfrac{Q^2}{2\pi\epsilon_0mR}}=\dfrac{Q}{\sqrt{2\pi\epsilon_0mR}} \).

Teacher's Note:
1. All the ring's charge is at the same distance R from the centre.
2. Potential energy of the system falls as the particle approaches.

Answer: (b) \( \dfrac{qQ}{4\pi\epsilon_0K} \)

Explanation:
1. At the closest approach all the kinetic energy has become potential energy.
2. \( K=\dfrac{qQ}{4\pi\epsilon_0r_0} \), so \( r_0=\dfrac{qQ}{4\pi\epsilon_0K} \).

Teacher's Note:
1. The nucleus is heavy and fixed, so it takes no kinetic energy.
2. Equate kinetic energy at infinity with potential energy at closest approach.