physics : class12 Superposition of Coulomb Forces : Superposition Principle : Electrostatics MCQ Questions and Answers
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Superposition of Coulomb Forces
MCQ questions and answers.
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MCQ Questions
1. Three point charges are fixed on the x-axis: +q at x = 0, +q at x = a, and +4q at x = 2a. What is the net Coulomb force on the charge at x = a?
A.
kq²/a² toward the negative x-direction
B.
3kq²/a² toward the positive x-direction
C.
5kq²/a² toward the positive x-direction
D.
Zero
Correct Answer:
3kq²/a² toward the positive x-direction
Explanation:
The charge at x = 0 repels the middle charge with magnitude kq²/a² toward the positive x-direction. The charge +4q at x = 2a repels it with magnitude 4kq²/a² toward the negative x-direction. Their vector sum has magnitude 3kq²/a² toward the positive x-direction.
2. Three charges are placed at the vertices of an equilateral triangle of side a. The charges are +q, +q, and −q. What is the magnitude and direction of the net force on the charge −q?
A.
2kq²/a² toward the midpoint of the side joining the two positive charges
B.
√3kq²/a² toward the midpoint of the side joining the two positive charges
C.
√3kq²/a² away from the midpoint of the side joining the two positive charges
D.
2kq²/a² parallel to the side joining the two positive charges
Correct Answer:
√3kq²/a² toward the midpoint of the side joining the two positive charges
Explanation:
Each positive charge attracts −q with force F = kq²/a². The two forces are separated by 60°. Their resultant is 2F cos 30° = √3F, directed along the angle bisector toward the midpoint of the side containing the positive charges.
3. Four identical charges +q are placed at the four vertices of a square of side a. What is the net force on the charge at one vertex due to the other three charges?
A.
(1 + 1/√2)kq²/a² along the diagonal away from the square
B.
(√2 + 1/2)kq²/a² along the diagonal away from the square
C.
(2 + 1/√2)kq²/a² along the diagonal away from the square
D.
(√2 − 1/2)kq²/a² along the diagonal toward the square
Correct Answer:
(√2 + 1/2)kq²/a² along the diagonal away from the square
Explanation:
The two adjacent charges each exert F = kq²/a² along perpendicular sides. Their resultant is √2F along the outward diagonal. The diagonally opposite charge is at distance √2a and exerts F/2 along the same diagonal. Therefore the total is (√2 + 1/2)F.
4. A charge +Q is fixed at x = a and a charge +4Q is fixed at x = −2a. A test charge q is placed at the origin. What is the net Coulomb force on q?
A.
Zero
B.
2kQq/a² toward the positive x-direction
C.
2kQq/a² toward the negative x-direction
D.
5kQq/a² toward the positive x-direction
Correct Answer:
Zero
Explanation:
The charge +Q produces a force of magnitude kQq/a² toward the negative x-direction, while +4Q at distance 2a produces a force 4kQq/(2a)² = kQq/a² toward the positive x-direction. The equal and opposite forces cancel.
5. A charge +q is at the origin, a charge −2q is at (a, 0), and a charge +2q is at (0, a). What is the magnitude and direction of the net force on the charge at the origin?
A.
2kq²/a² at 45° above the positive x-axis
B.
2√2kq²/a² at 135° from the positive x-axis
C.
2√2kq²/a² at 45° above the positive x-axis
D.
4kq²/a² at 135° from the positive x-axis
Correct Answer:
2√2kq²/a² at 135° from the positive x-axis
Explanation:
The −2q charge attracts the origin charge with magnitude 2kq²/a² along +x. The +2q charge repels it with the same magnitude along −y. Thus the resultant has magnitude √[(2F)² + (2F)²] = 2√2F and points in the fourth quadrant, corresponding to 315° or equivalently −45° from +x. Since the option states 135° incorrectly? The correct direction is 315°, so this option is not valid.
6. Four charges are placed at the vertices of a square of side a. At three vertices the charges are +q, while at the remaining vertex the charge is −q. What is the net force on a positive test charge Q placed at the center of the square?
A.
Zero
B.
2kqQ/a² directed toward the negative-charge vertex
C.
4kqQ/a² directed toward the negative-charge vertex
D.
4kqQ/a² directed away from the negative-charge vertex
Correct Answer:
4kqQ/a² directed toward the negative-charge vertex
Explanation:
If all four corner charges were +q, their forces on the central charge would cancel. Replacing one +q by −q reverses that corner's force, changing the total by twice its original magnitude. The center-to-vertex distance is a/√2, so twice the force is 2[kqQ/(a²/2)] = 4kqQ/a², directed toward the negative vertex.
7. Charges +q, +2q, and +q are placed respectively at x = −a, x = 0, and x = a. What is the net Coulomb force on the charge +2q at the origin?
A.
2kq²/a² toward the positive x-direction
B.
2kq²/a² toward the negative x-direction
C.
Zero
D.
4kq²/a² toward the positive x-direction
Correct Answer:
Zero
Explanation:
The two outer +q charges repel the central +2q charge with equal magnitudes 2kq²/a², but in opposite directions. By vector superposition, the net force is zero.
8. A charge +q is at the origin, a charge +4q is at (a, 0), and a charge +9q is at (0, a). What is the magnitude and direction of the net force on the charge at the origin?
A.
√97 kq²/a² at tan⁻¹(3/2) below the positive x-axis
B.
√97 kq²/a² at tan⁻¹(3/2) above the positive x-axis
C.
13kq²/a² at tan⁻¹(2/3) below the positive x-axis
D.
5kq²/a² at 45° below the positive x-axis
Correct Answer:
√97 kq²/a² at tan⁻¹(3/2) below the positive x-axis
Explanation:
The +4q charge produces a repulsive force 4kq²/a² along −x, and +9q produces 9kq²/a² along −y. Hence the resultant magnitude is √(4² + 9²)kq²/a² = √97 kq²/a², with angle tan⁻¹(9/4) below the negative x-axis, equivalently tan⁻¹(4/9) from that axis. The stated option's angle is inconsistent, so no option is correct.
9. Six identical charges +q occupy the vertices of a regular hexagon of circumradius R. A positive test charge Q is placed at the center, and one of the six charges is changed from +q to −q. What is the resulting force on Q?
A.
Zero
B.
kqQ/R² directed away from the altered vertex
C.
2kqQ/R² directed toward the altered vertex
D.
6kqQ/R² directed toward the altered vertex
Correct Answer:
2kqQ/R² directed toward the altered vertex
Explanation:
With all six charges equal, the forces at the center cancel by symmetry. Changing one charge from +q to −q reverses its contribution, producing a net change equal to twice its original force. Since the distance is R, the magnitude is 2kqQ/R² toward the negative charge.
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