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Positive and negative point charges of equal magnitude are kept at (0,0,a/2) and (0,0,-a/2), respectively. The work done by the electric field when another positive point charge in moved from (-a,0,0) to (0,a,0) is.
- a)negative
- b)depends on the path connecting the initial and final positions
- c)zero
- d)positive
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Positive and negative point charges of equal magnitude are kept at (0,...
Solution:
Concept:
The work done by an electric field on a charge is given by the formula:
W = q(Vf - Vi)
where,
q = charge
Vf = final potential energy
Vi = initial potential energy
Given:
Two point charges of equal magnitude are kept at (0,0,a/2) and (0,0,-a/2), respectively.
Another positive point charge is moved from (-a,0,0) to (0,a,0).
Explanation:
Let us assume that the charge being moved has a charge q.
The potential energy of the charge at position (-a,0,0) is given by:
V1 = kq(Q1)/r1
where,
Q1 = magnitude of the charge at (0,0,a/2)
r1 = distance between the charge being moved and the charge at (0,0,a/2)
Similarly, the potential energy of the charge at position (0,a,0) is given by:
V2 = kq(Q2)/r2
where,
Q2 = magnitude of the charge at (0,0,-a/2)
r2 = distance between the charge being moved and the charge at (0,0,-a/2)
The net potential energy change of the charge being moved is given by:
ΔV = V2 - V1
Since the magnitudes of the two charges are equal, the potential energies at the two positions will be equal in magnitude and opposite in sign.
Therefore, ΔV = 0, which means that the work done by the electric field is zero.
Hence, the correct answer is option (C).
Concept:
The work done by an electric field on a charge is given by the formula:
W = q(Vf - Vi)
where,
q = charge
Vf = final potential energy
Vi = initial potential energy
Given:
Two point charges of equal magnitude are kept at (0,0,a/2) and (0,0,-a/2), respectively.
Another positive point charge is moved from (-a,0,0) to (0,a,0).
Explanation:
Let us assume that the charge being moved has a charge q.
The potential energy of the charge at position (-a,0,0) is given by:
V1 = kq(Q1)/r1
where,
Q1 = magnitude of the charge at (0,0,a/2)
r1 = distance between the charge being moved and the charge at (0,0,a/2)
Similarly, the potential energy of the charge at position (0,a,0) is given by:
V2 = kq(Q2)/r2
where,
Q2 = magnitude of the charge at (0,0,-a/2)
r2 = distance between the charge being moved and the charge at (0,0,-a/2)
The net potential energy change of the charge being moved is given by:
ΔV = V2 - V1
Since the magnitudes of the two charges are equal, the potential energies at the two positions will be equal in magnitude and opposite in sign.
Therefore, ΔV = 0, which means that the work done by the electric field is zero.
Hence, the correct answer is option (C).
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Positive and negative point charges of equal magnitude are kept at (0,0,a/2) and (0,0,-a/2), respectively. The work done by the electric field when another positive point charge in moved from (-a,0,0) to (0,a,0) is.a)negativeb)depends on the path connecting the initial and final positionsc)zerod)positiveCorrect answer is option 'C'. Can you explain this answer?
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