If the electric field intensity between two charged parallel plates have the magnitude 2 × 102 N/C, then the magnitude of electrostatic force on an electron in the electric field will be A. 8.0 × 10^–22 N B. 3.2 × 10^–17 N C. 2.0 × 10^2 N D. 1.2 × 10^21 N
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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
=4πϵ01r2Q1Q2
=10×2×3
=60
F=7QE=760 (due to parallel plate capacitor)
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