physics : class12 Electric Field Due to Collinear Charges : Superposition Principle : Electrostatics MCQ Questions and Answers
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Electric Field Due to Collinear Charges
MCQ questions and answers.
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MCQ Questions
1. Two point charges +4q and −q are fixed at x = 0 and x = d, respectively. At which position on the x-axis is the net electric field zero?
A.
x = −2d
B.
x = d/2
C.
x = 2d
D.
No finite point exists
Correct Answer:
x = 2d
Explanation:
For x > d, the fields due to +4q and −q are opposite. Equating their magnitudes, 4/x² = 1/(x − d)², gives 2/x = 1/(x − d), hence x = 2d.
2. Equal positive charges +q and +4q are placed at x = 0 and x = d, respectively. Where between them is the resultant electric field zero?
A.
x = d/4
B.
x = d/3
C.
x = d/2
D.
x = 2d/3
Correct Answer:
x = d/3
Explanation:
Between like charges, the two fields oppose each other. At x, q/x² = 4q/(d − x)². Therefore d − x = 2x, giving x = d/3.
3. Charges +q and −4q are fixed at x = 0 and x = d, respectively. The net electric field vanishes at which point on the x-axis?
A.
x = −d
B.
x = d/3
C.
x = 2d
D.
No finite point exists
Correct Answer:
x = −d
Explanation:
The zero-field point must lie outside the interval, on the side of the smaller-magnitude charge. For x < 0, equating magnitudes gives q/x² = 4q/(d − x)², which yields x = −d.
4. Charges +q, −4q, and +q are placed at x = −a, 0, and +a, respectively. What is the electric field at x = 2a?
A.
kq/(9a²) toward +x
B.
kq/(9a²) toward −x
C.
kq/a² toward +x
D.
5kq/(9a²) toward +x
Correct Answer:
kq/(9a²) toward +x
Explanation:
At x = 2a, the fields from the charges at −a and +a point toward +x, with magnitudes kq/(3a)² and kq/a². The −4q charge produces a field toward −x of magnitude 4kq/(2a)² = kq/a². Thus the net field is kq/(9a²) toward +x.
5. A charge +q is placed at x = −a and a charge −4q at x = +a. What is the resultant electric field at the origin?
A.
3kq/a² toward +x
B.
5kq/a² toward +x
C.
3kq/a² toward −x
D.
5kq/a² toward −x
Correct Answer:
5kq/a² toward +x
Explanation:
At the origin, +q at −a produces a field kq/a² toward +x. The −4q at +a also produces a field toward +x, with magnitude 4kq/a². Their sum is 5kq/a² toward +x.
6. Two charges +q and +9q are fixed at x = 0 and x = d. What is the resultant electric field at the midpoint x = d/2?
A.
32kq/d² toward +x
B.
32kq/d² toward −x
C.
16kq/d² toward +x
D.
Zero
Correct Answer:
32kq/d² toward −x
Explanation:
At the midpoint, each charge is at distance d/2. The field due to +q is toward +x and equals 4kq/d², while the field due to +9q is toward −x and equals 36kq/d². The net field is 32kq/d² toward −x.
7. Charges +Q and +4Q are placed at x = 0 and x = 3a, respectively. At what point between them is the net electric field zero?
A.
x = a/3
B.
x = a
C.
x = 2a
D.
x = 3a/2
Correct Answer:
x = a
Explanation:
For a point x between the charges, Q/x² = 4Q/(3a − x)². Taking positive square roots gives 3a − x = 2x, so x = a.
8. Charges +q, −2q, and +q are placed at x = −d, 0, and +d, respectively. What is the net electric field at x = 2d?
A.
kq/(18d²) toward +x
B.
11kq/(18d²) toward +x
C.
11kq/(18d²) toward −x
D.
kq/(2d²) toward +x
Correct Answer:
11kq/(18d²) toward +x
Explanation:
The fields from the charges at −d and +d point toward +x and contribute kq/(3d)² and kq/d². The −2q charge contributes toward −x with magnitude 2kq/(2d)² = kq/(2d²). Hence E = kq/d²(1/9 + 1 − 1/2) = 11kq/(18d²) toward +x.
9. A charge +Q is fixed at x = 0 and an unknown positive charge q is fixed at x = d. If the electric field is zero at x = 2d, what is q?
A.
Q/16
B.
Q/9
C.
Q/4
D.
4Q
Correct Answer:
Q/4
Explanation:
At x = 2d, the distances from +Q and q are 2d and d, respectively. Their opposite-directed fields must satisfy kQ/(2d)² = kq/d², giving q = Q/4.
10. Charges +2q and −8q are placed at x = 0 and x = 3d, respectively. Where on the x-axis is the net electric field zero?
A.
x = −3d
B.
x = d
C.
x = 3d/2
D.
x = 6d
Correct Answer:
x = −3d
Explanation:
For opposite charges, the zero-field point lies outside on the side of the smaller charge, so let x = −u. Equating magnitudes gives 2q/u² = 8q/(3d + u)². Thus 3d + u = 2u, so u = 3d and x = −3d.
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