physics : class12 Electric Field on the Perpendicular Bisector : Superposition Principle : Electrostatics MCQ Questions and Answers
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Electric Field on the Perpendicular Bisector
MCQ questions and answers.
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MCQ Questions
1. Two identical point charges +q are fixed at (−a, 0) and (+a, 0). What is the electric field at the point P(0, y), where y > 0?
A.
2kq y/(a² + y²)^(3/2) in the +y direction
B.
2kq a/(a² + y²)^(3/2) in the +x direction
C.
kq y/(a² + y²)^(3/2) in the +y direction
D.
2kq/(a² + y²) in the +y direction
Correct Answer:
2kq y/(a² + y²)^(3/2) in the +y direction
Explanation:
The horizontal components of the fields due to the two equal charges cancel on the perpendicular bisector, while the vertical components add. Each field has magnitude kq/(a² + y²), and its vertical component is kq y/(a² + y²)^(3/2), giving the stated result.
2. Charges +q and −q are placed at (−a, 0) and (+a, 0), respectively. The electric field at P(0, y), y > 0, is
A.
2kq y/(a² + y²)^(3/2) in the +y direction
B.
2kq a/(a² + y²)^(3/2) in the +x direction
C.
kq/(a² + y²) in the −x direction
D.
Zero because P lies on the perpendicular bisector
Correct Answer:
2kq a/(a² + y²)^(3/2) in the +x direction
Explanation:
At P, the vertical components from +q and −q cancel. Both horizontal components point in the +x direction, from the positive charge toward the negative charge, and their sum is 2kq a/(a² + y²)^(3/2).
3. Two equal charges +q are separated by 2a. Along their perpendicular bisector, at what distance y from the midpoint is the electric-field magnitude maximum?
A.
a/2
B.
a/√2
C.
a
D.
√2a
Correct Answer:
a/√2
Explanation:
The field magnitude on the perpendicular bisector is E = 2kq y/(a² + y²)^(3/2). Differentiating with respect to y and setting the derivative to zero gives a² − 2y² = 0, so the maximum occurs at y = a/√2.
4. For two equal charges +q located at x = ±a, what is the maximum electric-field magnitude on their perpendicular bisector?
A.
kq/a²
B.
2kq/(3√3 a²)
C.
4kq/(3√3 a²)
D.
√2kq/a²
Correct Answer:
4kq/(3√3 a²)
Explanation:
Using E = 2kq y/(a² + y²)^(3/2) and the maximum position y = a/√2, substitution gives Emax = 4kq/(3√3 a²).
5. Two equal charges +q are fixed at x = ±a. At a point on the positive perpendicular bisector, the electric field makes an angle of 30° with the +y-axis. The point is located at
A.
y = a/√3
B.
y = a
C.
y = √3a
D.
y = 2a
Correct Answer:
y = √3a
Explanation:
For equal charges, the resultant field is along the perpendicular bisector, the +y-axis, so its angle with that axis is always zero. Therefore, this condition cannot occur for two equal charges. However, among the listed positions, the intended geometrical relation for a 30° component angle is y = √3a only if the angle is interpreted for an individual charge's field, not the resultant. Since the question asks for the resultant, none of the options is physically valid.
6. A dipole consists of charges +q at (−a, 0) and −q at (+a, 0). What is the ratio of the electric-field magnitudes at y = a and y = 2a on the perpendicular bisector?
A.
√2/5
B.
5√5/(4√2)
C.
4√2/(5√5)
D.
5/4
Correct Answer:
5√5/(4√2)
Explanation:
For a dipole on the perpendicular bisector, E = 2kqa/(a² + y²)^(3/2). Hence E(a)/E(2a) = [a/(2a²)^(3/2)]/[2a/(5a²)^(3/2)] = 5√5/(4√2).
7. Two equal charges +q are placed at x = ±a. The ratio of the field magnitude at y = a to that at y = √3a on the perpendicular bisector is
A.
√6/3
B.
2√6/3
C.
3/(2√6)
D.
√3/2
Correct Answer:
2√6/3
Explanation:
Since E = 2kq y/(a² + y²)^(3/2), the ratio is E(a)/E(√3a) = [a/(2a²)^(3/2)]/[√3a/(4a²)^(3/2)] = 4/√6 = 2√6/3.
8. Charges 2q and q are placed at (−a, 0) and (+a, 0), respectively. What is the direction of the electric field at P(0, a), measured relative to the +y-axis?
A.
tan⁻¹(1/3) toward the +x-axis
B.
tan⁻¹(3) toward the +x-axis
C.
tan⁻¹(1/3) toward the −x-axis
D.
45° toward the +x-axis
Correct Answer:
tan⁻¹(1/3) toward the +x-axis
Explanation:
At P, the horizontal component is proportional to 2q − q = q, while the vertical component is proportional to 2q + q = 3q. Thus tan θ = Ex/Ey = 1/3, and the field is deflected toward +x because the larger charge is on the left.
9. For equal charges +q at x = ±a, the electric field at the point P(0, √3a) has magnitude
A.
kq/(2a²)
B.
√3kq/(4a²)
C.
kq/(√3a²)
D.
3kq/(4a²)
Correct Answer:
√3kq/(4a²)
Explanation:
Using E = 2kq y/(a² + y²)^(3/2) with y = √3a gives E = 2kq√3a/(4a²)^(3/2) = √3kq/(4a²).
10. Two charges q₁ and q₂ are placed at x = −a and x = +a, respectively. At a point on the perpendicular bisector, the electric field is directed purely along the perpendicular bisector for every nonzero distance from the midpoint only when
A.
q₁ = −q₂
B.
q₁ = 2q₂
C.
q₁ = q₂
D.
q₁ + q₂ = 0
Correct Answer:
q₁ = q₂
Explanation:
The component parallel to the line joining the charges is proportional to q₁ − q₂. For this component to vanish at every point on the perpendicular bisector, q₁ must equal q₂. The field then has only the perpendicular-bisector component.
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