A capacitor having C=10μF; Potential across plates, V = 10V; Distance between plates 10mm; And area of plates 10cm2, has air between the plates. What is the force on one of the plates.
A
4.65×105N
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B
5.65×105N
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C
6.65×105N
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D
None of these
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Solution
The correct option is B5.65×105N Let’s first find the charge on the plate,
Q = CV Q=10×10−6×10 Q=10−4C Now we know that the force will simply be →F=→EQ And since E is approximately constant between the plates and is given by E=σ2ε0 E=Q2Aε0 F=Q22Aε0 F=(10−4)22×10×10−4×8.854×10−12 F=5.65×105N