physics : class12 Electric Field Due to Noncollinear Charges : Superposition Principle : Electrostatics MCQ Questions and Answers
Practice important
Electric Field Due to Noncollinear Charges
MCQ questions and answers.
Page 1
of
1
|
10
MCQ Questions
1. A charge +q is placed at the origin and an identical charge +q at (a, 0). What is the electric field at the point (0, a), where k = 1/(4πε₀)?
A.
(kq/a²)[-1/(2√2) i + (1 + 1/(2√2)) j]
B.
(kq/a²)[1/(2√2) i + (1 - 1/(2√2)) j]
C.
(kq/a²)[-1/(2√2) i + (1 - 1/(2√2)) j]
D.
(kq/a²)[1/(2√2) i + (1 + 1/(2√2)) j]
Correct Answer:
(kq/a²)[-1/(2√2) i + (1 + 1/(2√2)) j]
Explanation:
The field due to the charge at the origin is (kq/a²)j. The displacement from (a, 0) to (0, a) is -a i + a j, whose magnitude is √2a; hence its field is (kq/a²)[-1/(2√2)i + 1/(2√2)j]. Adding the two vectors gives the stated result.
2. Three charges are placed at the vertices of an equilateral triangle of circumradius R. The charges at two vertices are +q and the charge at the third vertex is -q. What is the electric field at the triangle's center, directed relative to the radius from the center toward the -q vertex?
A.
Zero
B.
√3kq/R² directed away from the -q vertex
C.
2kq/R² directed toward the -q vertex
D.
2kq/R² directed away from the -q vertex
Correct Answer:
2kq/R² directed toward the -q vertex
Explanation:
At the center, each charge produces a field of magnitude kq/R². Using the vector sum of the three radial unit vectors, the combined field of the two +q charges and the -q charge is 2kq/R² in the direction of the radius pointing toward the -q vertex.
3. A charge +2q is at (0, 0), and a charge -q is at (a, 0). What is the electric field at (0, a)?
A.
(kq/a²)[1/(2√2) i + (2 + 1/(2√2)) j]
B.
(kq/a²)[1/(2√2) i + (2 - 1/(2√2)) j]
C.
(kq/a²)[-1/(2√2) i + (2 - 1/(2√2)) j]
D.
(kq/a²)[-1/(2√2) i + (2 + 1/(2√2)) j]
Correct Answer:
(kq/a²)[1/(2√2) i + (2 - 1/(2√2)) j]
Explanation:
The +2q charge produces (2kq/a²)j. From (a, 0) to (0, a), the displacement is -a i + a j. Since the source charge is negative, its field points toward (a, 0), giving (kq/a²)[1/(2√2)i - 1/(2√2)j]. Their vector sum is the stated expression.
4. Two equal charges +q are located at (a, 0) and (0, a). What is the magnitude and direction of the electric field at the origin?
A.
√2kq/a², directed along (i + j)
B.
2kq/a², directed along (i + j)
C.
√2kq/a², directed along -(i + j)
D.
kq/a², directed along (i - j)
Correct Answer:
√2kq/a², directed along -(i + j)
Explanation:
Each charge is at distance a from the origin. The field due to the charge at (a, 0) is -(kq/a²)i, while that due to the charge at (0, a) is -(kq/a²)j. Their resultant is -(kq/a²)(i + j), with magnitude √2kq/a².
5. Charges +q and +4q are placed at the points (0, 0) and (3a, 4a), respectively. At what point between them is the net electric field zero?
A.
(3a/5, 4a/5)
B.
(a, 4a/3)
C.
(2a, 8a/3)
D.
(3a/2, 2a)
Correct Answer:
(a, 4a/3)
Explanation:
The charges are separated by 5a. For cancellation between like charges, q/r₁² = 4q/r₂², so r₂ = 2r₁. Since r₁ + r₂ = 5a, r₁ = 5a/3. Moving this distance from the origin along the direction (3/5)i + (4/5)j gives (a, 4a/3).
6. A charge +q is placed at (a, 0), and a charge -q at (0, a). What is the electric field at the origin?
A.
(kq/a²)(i + j)
B.
(kq/a²)(-i + j)
C.
(kq/a²)(i - j)
D.
(kq/a²)(-i - j)
Correct Answer:
(kq/a²)(-i + j)
Explanation:
The positive charge at (a, 0) produces a field toward the negative x-direction, namely -(kq/a²)i. The negative charge at (0, a) attracts a positive test charge upward, producing +(kq/a²)j. Superposition gives (kq/a²)(-i + j).
7. Charges +q, -q, and +2q are placed at (a, 0), (0, a), and (a, a), respectively. What is the electric field at the origin?
Explanation:
The fields due to +q at (a, 0) and -q at (0, a) are -(kq/a²)i and +(kq/a²)j, respectively. The distance to +2q at (a, a) is √2a, so its field at the origin is -(kq/a²)(i + j)/√2. Adding all three vectors yields the stated result.
8. A charge +q is at (a, 0), a charge +2q is at (0, 2a), and a charge -2q is at (-a, 0). What is the electric field at the origin?
A.
(kq/a²)(-3i - 1/2j)
B.
(kq/a²)(-i - 3/2j)
C.
(kq/a²)(-3i + 1/2j)
D.
(kq/a²)(3i - 1/2j)
Correct Answer:
(kq/a²)(-3i - 1/2j)
Explanation:
The +q at (a, 0) gives -(kq/a²)i. The +2q at (0, 2a) gives -(kq/2a²)j. The -2q at (-a, 0) gives -(2kq/a²)i because the field points toward the negative charge. The resultant is (kq/a²)(-3i - 1/2j).
9. A charge +q is at (0, 0), and a charge -q is at (a, 0). What is the electric field at (a, a)?
Explanation:
The field from +q at the origin has displacement a i + a j and equals (kq/a²)(i + j)/(2√2). The field from -q at (a, 0) points downward and equals -(kq/a²)j. Adding these gives the stated vector.
10. Four charges are placed at the vertices of a square of circumradius R. Charges +q are at two adjacent vertices, while charges -q are at the two opposite adjacent vertices. What is the electric field at the square's center?
A.
2√2kq/R² directed toward the midpoint of the +q side
B.
2kq/R² directed toward the midpoint of the -q side
C.
2√2kq/R² directed toward the midpoint of the -q side
D.
Zero, because the square is symmetric
Correct Answer:
2√2kq/R² directed toward the midpoint of the -q side
Explanation:
Each charge produces a field of magnitude kq/R² at the center. The fields from the two +q charges point away from them, and those from the two -q charges point toward them. Resolving the four vectors along the square's symmetry axis gives a resultant magnitude 2√2kq/R² directed toward the side containing the -q charges.
Short Questions and Answers
Practice short questions and answers related to
Electric Field Due to Noncollinear Charges
.
Short questions are currently unavailable for this page.