JEE Physics Electric Charges and Fields MCQs Set 20

Answer: (a) 180°

Explanation:
1. With no gravity the only forces on each ball are the string tension and the mutual repulsion.
2. Equilibrium needs these two to be equal and opposite, so the repulsion must act along the string.
3. The balls must therefore lie on a straight line through the hook with the strings pointing in opposite directions.

Teacher's Note:
1. Gravity is what tilts the strings toward each other on Earth; remove it and the strings align with the line joining the charges.
2. The angle then equals 180 degrees.

Answer: (d) 8.94 × 10⁻² m

Explanation:
1. In a medium the force is \( F=\dfrac{1}{4\pi\epsilon_0 K}\dfrac{q_1q_2}{r^2} \).
2. Equal force requires \( r'^2=r^2/K \), so \( r'=r/\sqrt K=20/\sqrt5\ \text{cm} \).
3. This is 8.94 cm, that is \( 8.94\times10^{-2}\ \text{m} \).

Teacher's Note:
1. Dividing the force by K means the distance must shrink by the square root of K to compensate.
2. Convert centimetres to metres before comparing options.

Answer: (d) 2

Explanation:
1. \( F\propto q(Q-q) \) for a fixed separation.
2. Setting \( \dfrac{d}{dq}[q(Q-q)]=Q-2q=0 \) gives \( q=Q/2 \).
3. Hence \( Q/q=2 \).

Teacher's Note:
1. A product of two quantities with a fixed sum is greatest when the two are equal.
2. This is the same idea as maximum area for a fixed perimeter.

Answer: (b) 4500 V/m

Explanation:
1. Displacement from the charge: \( (8-2)\hat i+(-5-3)\hat j=6\hat i-8\hat j \), so \( r=10\ \text{m} \).
2. \( E=\dfrac{9\times10^9\times50\times10^{-6}}{10^2}=4500\ \text{V/m} \).

Teacher's Note:
1. The field depends only on the separation from the charge, not on the coordinates themselves.
2. Use micro = 10⁻⁶.

Answer: (c) both (i) and (ii) are correct

Explanation:
1. The field is radial, so \( \vec E\parallel\vec r \). For (i): \( \vec r_A\cdot\vec r_B=1+2-3=0 \), so the fields are perpendicular.
2. For (ii): \( r_B^2=3 \) and \( r_C^2=12 \); since \( E\propto1/r^2 \), \( E_B/E_C=12/3=4 \).
3. Both statements are correct.

Teacher's Note:
1. The field of a point charge points along the position vector, so a dot product of positions tests perpendicularity.
2. Inverse-square dependence converts a ratio of squared distances into a ratio of fields.

Answer: (b) \( x^{-2} \)

Explanation:
1. \( E=\dfrac{kQx}{(x^2+R^2)^{3/2}} \).
2. For \( x\gg R \), \( E\approx kQ/x^2 \).

Teacher's Note:
1. Far from any finite charge distribution it looks like a point charge.
2. Hence the inverse-square law holds.

Answer: (c) directed towards the gap and proportional to \( 1/R^3 \)

Explanation:
1. The full ring gives zero field at the centre, so the remaining ring gives the negative of the field of the removed piece.
2. The removed charge is \( q=\dfrac{Qd}{2\pi R} \), giving \( E=\dfrac{kq}{R^2}=\dfrac{kQd}{2\pi R^3} \).
3. The gap's own field points away from the gap, so the remaining ring's field points towards the gap.

Teacher's Note:
1. Superposition: ring minus gap equals zero minus the gap's field.
2. Charge per length is Q divided by the circumference.

Answer: (c) higher potential and lower potential energy

Explanation:
1. A free charge moves so that its potential energy decreases.
2. For a negative charge \( U=qV \) with \( q<0 \), so U falls when V rises.
3. It therefore moves towards higher potential and lower potential energy.

Teacher's Note:
1. Positive charges go to lower potential, negative charges to higher potential.
2. Both lose potential energy.

Answer: (d) 1 : 4

Explanation:
1. \( U=k\left[\dfrac{-qQ}{r}+\dfrac{-qQ}{r}+\dfrac{q^2}{2r}\right]=0 \).
2. So \( -2Q+q/2=0 \), giving \( Q/q=1/4 \).

Teacher's Note:
1. The outer charges are 2r apart.
2. Sum all three pair energies.

Answer: (c) X will stop and Y will move with velocity u

Explanation:
1. After a long time the particles are far apart again, so the potential energy is zero as at the start.
2. Momentum and kinetic energy are both conserved, as in an elastic collision.
3. Equal masses exchange velocities, so X stops and Y moves with u.

Teacher's Note:
1. This is an elastic head-on collision carried out by a repulsive force.
2. Equal masses simply swap velocities.

Answer: (c) \( \dfrac{-2q}{2+\sqrt2} \)

Explanation:
1. Pair distances: Q to the near +q is a, Q to the far +q is \( a\sqrt2 \), and the two +q are a apart.
2. \( U=\dfrac{k}{a}\left[q^2+Qq+\dfrac{Qq}{\sqrt2}\right]=0 \) gives \( Q=-\dfrac{q}{1+1/\sqrt2} \).
3. This simplifies to \( -\dfrac{2q}{2+\sqrt2}=-(2-\sqrt2)q \).

Teacher's Note:
1. Identify which pair is the hypotenuse before writing the sum.
2. Rationalise to compare with the options.

Answer: (b) 10 V

Explanation:
1. \( E=F/q=3000/3=1000\ \text{N/C} \).
2. \( \Delta V=Ed=1000\times0.01=10\ \text{V} \).

Teacher's Note:
1. Convert 1 cm to 0.01 m.
2. The potential difference equals field times separation along the field.

Answer: (b) −1.52 × 10⁵ V

Explanation:
1. O is 0.5 m from B and from C and \( \sqrt3/2=0.866\ \text{m} \) from A.
2. \( V=9\times10^9\left[\dfrac{-6\times10^{-6}}{0.866}+\dfrac{-2\times10^{-6}}{0.5}+\dfrac{-3\times10^{-6}}{0.5}\right] \).
3. This equals \( 9\times10^9\times(-16.93\times10^{-6})=-1.52\times10^5\ \text{V} \).

Teacher's Note:
1. Potential is a scalar; add the signed contributions.
2. Distance AO is the height of the triangle.

Answer: (b) 0

Explanation:
1. The field direction is along \( (\hat i+\hat j)/\sqrt2 \) and the displacement from A to B is \( (-2,2) \).
2. \( \Delta V=-\vec E\cdot\vec d=-\dfrac{100}{\sqrt2}(-2+2)=0 \).

Teacher's Note:
1. The line y = 3 + x has slope 1, so the field makes 45° with the x-axis.
2. The two points lie on the same equipotential, which is perpendicular to the field.

Answer: (c) \( z=\dfrac{\text{constant}}{x^4y^2} \)

Explanation:
1. \( E_x=-\partial V/\partial x=4axy\sqrt z \) gives \( V=-2ax^2y\sqrt z+f(y,z) \).
2. This satisfies \( E_y=2ax^2\sqrt z \) and \( E_z=ax^2y/\sqrt z \), so \( V=-2ax^2y\sqrt z \).
3. Constant V means \( x^2y\sqrt z=c \), i.e. \( z=c'/(x^4y^2) \).

Teacher's Note:
1. Integrate one component and check the other two.
2. Squaring \( x^2y\sqrt z=c \) gives the form with x⁴y².

Answer: (a) 2 : 1

Explanation:
1. Axial: \( E=\dfrac{2kp}{r^3} \); equatorial: \( E=\dfrac{kp}{r^3} \).
2. The ratio is 2 : 1.

Teacher's Note:
1. The axial field is twice the equatorial field at equal distance.
2. The equatorial field also points opposite to the dipole moment.

Answer: (c) \( 2qa\,\hat j \)

Explanation:
1. The total charge is \( -2q-2q+3q+q=0 \), so the dipole moment is origin-independent.
2. \( \vec p=\sum q_i\vec r_i=-2q(-a\hat i)-2q(a\hat i)+3q(a\hat j)+q(-a\hat j) \).
3. The x-parts cancel and the y-part is \( 2qa \).

Teacher's Note:
1. A neutral system has a well-defined dipole moment.
2. Compute component by component.

Answer: (b) 0.1 rad/s

Explanation:
1. \( p=qd=10^{-6}\times3=3\times10^{-6}\ \text{C m} \) and \( pE=3\times10^{-6}\times2\times10^4=0.06\ \text{N m} \).
2. Rotation is about the centre of mass; the moment of inertia is \( I=\mu d^2=\dfrac{1\times2}{1+2}\times9=6\ \text{kg m}^2 \).
3. \( \omega=\sqrt{pE/I}=\sqrt{0.06/6}=0.1\ \text{rad/s} \).

Teacher's Note:
1. For two masses, the moment of inertia about the centre of mass is the reduced mass times the square of the separation.
2. Small-angle torque is pE times theta.

Answer: (b) on the line AB, outside AB, 3 m beyond B

Explanation:
1. For unlike charges the null point lies outside the segment, on the side of the smaller charge.
2. Let it be x from A: \( \dfrac{4Q}{x^2}=\dfrac{Q}{(x-3)^2} \), so \( 2(x-3)=x \) and x = 6 m.
3. This is 3 m beyond B.

Teacher's Note:
1. Between unlike charges the fields add, so no null point is there.
2. Take the square root of the charge ratio for the distance ratio.

Answer: (a) \( Q=-\dfrac49q \text{ at a distance } l/3 \text{ from } q \)

Explanation:
1. Q must lie between the charges and be negative to attract both.
2. Let it be x from q. Equilibrium of q: \( Q=4qx^2/l^2 \); of 4q: \( Q=q(l-x)^2/l^2 \).
3. Equating: \( 2x=l-x \), x = l/3, and \( |Q|=4q/9 \).

Teacher's Note:
1. Write the force balance for the two outer charges separately.
2. The third charge's own equilibrium is then automatic.

Answer: (d) 18 V

Explanation:
1. The displacement is \( (-3,-4,0) \) and \( r=5\ \text{m} \).
2. \( V=\dfrac{9\times10^9\times10^{-8}}{5}=18\ \text{V} \).

Teacher's Note:
1. Only the distance matters for potential.
2. The field there has no z-component because both points have z = 2.

Answer: (b) execute oscillatory but not simple harmonic motion

Explanation:
1. The net force on Q is always directed towards the origin along the x-axis, so Q oscillates about the origin.
2. The restoring force is proportional to x only for \( x\ll a \).
3. The amplitude 2a is not small, so the motion is oscillatory but not SHM.

Teacher's Note:
1. Compare the amplitude with the separation of the fixed charges.
2. Q passes through the origin with maximum speed and continues to −2a.

Answer: (d) 0.5 m/s

Explanation:
1. \( a=\dfrac{qE}{m}=\dfrac{10^{-6}\times300}{10^{-3}}=0.3\ \text{m/s}^2 \), so the change in velocity has magnitude \( \Delta v=at=3\ \text{m/s} \).
2. The final speed lies between \( 4-3=1 \) and \( 4+3=7 \) m/s depending on direction.
3. 0.5 m/s is outside this range.

Teacher's Note:
1. The velocity change is a vector of fixed magnitude but arbitrary direction.
2. The final speed range comes from the triangle inequality.