Find the electric field at the center of the square of Fig. 26-26. Assume that q = 11.8 nC and a = 5.20 cm.
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Answer:
Let the distance between each corner and the center of square be x.
∴ s
2
=x
2
+x
2
⟹x=
2
s
Potential at the center due to charge at corner 1, V
1
=
x
kq
=
s
2
kq
The square is symmetric about its center.
Thus total potential at the center V=4V
1
=k
s
4
2
q
solution
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Explanation:
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