Charges +q and -q are placed at points A and B respectively which are a distance 2L apart, C is the midpoint between A and B. The work done is moving a charge +Q along the semicircle CRD is : ___.
Answers
Answer:
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Explanation:
from fig, AC = L,BC = L, BD = BC = L
AD = AB + BD
2L + L = 3L
Work done is equal to charge in potential energy.
In case-1, when charge +Q is situated at C
The electric potential energy of the system.
U1= k(q)(−q) + k(−q)(Q) + kqQ
2L L L
In case-2, when charge +Q is moved from C to D
The electric potential energy of the system in that case
U2 = k(q)(−q) + kqQ+ k(−q)(Q)
2L 3L L
Work done ΔU=U2 − U1
ΔU = kkq −kqQ
3L L
1 -2qQ
ΔU = 4πϵ0 3L
ΔU = -qQ
6πϵ0l
potential at c is 0 because the charges are equal and opposite and the distance are the same potential at d due to -q is greater than that A (+q) because D is closer to B therefore it is negative
