(A) Two charges q = 7.0 C are fixed in space a distance d = 5.0 cm apart, as sho
ID: 2137472 • Letter: #
Question
(A) Two charges q = 7.0 C are fixed in space a distance d = 5.0 cm apart, as shown in the figure. With V = 0 at infinity, what is the electric potential at point C?
(B) You bring a third charge q = 7.0C from infinity to C. How much work must you do?
(C) What is the potential energy U of the three-charge configuration when the third charge is in place?
Explanation / Answer
a.)Let the potential at C be Vc
Vc=2kq/a
a=d/2*sqrt(2)=d/dqrt(2).
so Vc=2kq*sqrt(2)/d
so Vc=3563818.18 V
b.) the work needed W = q(V-V0)=qVc=24.94 J
c.) The potential energy that is needed is increase by 24.94 J
the initial potential energy W=kq^2/d=8.82 J
so the potentail energy W'=W+?W=33.76 J
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