JEE Physics Electric Charges and Fields MCQs Set 17

Answer: (a) P-3, Q-1, R-4, S-2

Explanation:
1. P: the x components cancel by symmetry and all charges repel q upward, so the force is along +y.
2. Q: positive charges on the left push q toward +x and negative charges on the right attract it toward +x, so the force is along +x.
3. R: x components cancel; the nearer negative charges Q2 and Q3 attract q downward more strongly than Q1 and Q4 push it up, so the force is along -y.
4. S: y components cancel; nearer Q2 (negative) attracts toward -x and nearer Q3 (positive) pushes toward -x, beating the far charges, so the force is along -x.

Teacher's Note:
1. Apply Gauss law or symmetry first, then check each statement or option separately.
2. Verify the result with a limiting case.

Answer: (b) 6

Explanation:
1. Viewed along the line, the rectangle subtends a half angle \( \theta \) with \( \tan\theta=\frac{a/2}{\sqrt3a/2}=\frac1{\sqrt3} \), so \( \theta=30^\circ \) and the full angle is \( 60^\circ \).
2. The flux through length L of a full cylinder is \( \frac{\lambda L}{\epsilon_0} \), so the share through the rectangle is \( \frac{60}{360}\frac{\lambda L}{\epsilon_0}=\frac{\lambda L}{6\epsilon_0} \).
3. Hence n = 6.

Teacher's Note:
1. Apply Gauss law or symmetry first, then check each statement or option separately.
2. Verify the result with a limiting case.

Answer: (d) It is uniform, with magnitude and direction depending on the vector from O to P (not on the position inside the cavity)

Explanation:
1. Treat the cavity as the full sphere of density \( \rho \) plus a sphere of density \( -\rho \) at P.
2. Inside a uniformly charged sphere \( \vec E=\frac{\rho\vec r}{3\epsilon_0} \); the difference gives \( \vec E=\frac{\rho}{3\epsilon_0}(\vec r_O-\vec r_P)=\frac{\rho\vec a}{3\epsilon_0} \).
3. So the field is uniform, parallel to OP, and its magnitude depends on a but not on R2 or on the point chosen in the cavity.

Teacher's Note:
1. Apply Gauss law or symmetry first, then check each statement or option separately.
2. Verify the result with a limiting case.

Answer: (a) 3 and 4 only

Explanation:
1. The charge is outside the closed surface, so the total flux is zero (statement 1 false).
2. The flat surface is at varying distance from Q, so the normal component varies (statement 2 false).
3. Every point of the circumference is at distance \( \sqrt2R \) from Q, so it is an equipotential (3 true).
4. Flux through the flat disk is \( \frac{Q}{2\epsilon_0}(1-\cos\theta) \) with \( \cos\theta=\frac1{\sqrt2} \); the curved surface carries the negative of this (4 true).

Teacher's Note:
1. Apply Gauss law or symmetry first, then check each statement or option separately.
2. Verify the result with a limiting case.

Answer: (d) P-5, Q-3, R-1 and 4, S-2

Explanation:
1. Infinite plane: \( E=\frac{\sigma}{2\epsilon_0} \), independent of d (P-5). Infinite line: \( E\propto\frac1d \) (Q-3).
2. Point charge: \( E\propto\frac1{d^2} \) (R-1); two opposite line charges form a line dipole whose field also falls as \( \frac1{d^2} \) (R-4).
3. A point dipole gives \( E\propto\frac1{d^3} \) (S-2).

Teacher's Note:
1. Apply Gauss law or symmetry first, then check each statement or option separately.
2. Verify the result with a limiting case.

Answer: (a) E = 0 for r < R, rises abruptly to its maximum value at r = R, then falls as 1/r squared

Explanation:
1. By Gauss law, a closed surface inside the shell encloses no charge, so E = 0 for r < R.
2. Outside, \( E=\frac{Q}{4\pi\epsilon_0r^2} \), which has its largest value \( \frac{Q}{4\pi\epsilon_0R^2} \) just outside r = R.
3. So the curve is zero, jumps at R, then decays as inverse square.

Teacher's Note:
1. Apply Gauss law or symmetry first, then check each statement or option separately.
2. Verify the result with a limiting case.

Answer: (b) \( -2\sqrt2 \)

Explanation:
1. Let the side be a. The two charges q each exert \( \frac{kqQ}{a^2} \) at right angles, resultant \( \frac{\sqrt2kqQ}{a^2} \).
2. The other Q at the diagonal exerts \( \frac{kQ^2}{2a^2} \) in the opposite sense when q is of opposite sign.
3. Balance: \( \frac{Q^2}{2}=-\sqrt2qQ \), so \( \frac Qq=-2\sqrt2 \).

Teacher's Note:
1. Apply Gauss law or symmetry first, then check each statement or option separately.
2. Verify the result with a limiting case.