if electric field between plates of parallel plate capacitor is 2 N/C and charge on two plates are 10C and 3C then force on one of the plates is
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Explanation:
Capacitor =2N/C
Charge =10C and 3C
the magnitude of the force exerted by one plate on the other should have been
F=QE
=
4πϵ
0
1
r
2
Q
1
Q
2
=10×2×3
=60
F=
7
QE
=
7
60
(due to parallel plate capacitor)
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