Electrostatics Question 40

Question: Force of attraction between the plates of a parallel plate capacitor is [AFMC 1998]

Options:

A) q22ε0AK

B) q2ε0AK

C) q2ε0A

D) q22ε0A2K

Show Answer

Answer:

Correct Answer: A

Solution:

Force on one plate due to another is F = qE = q×σ2ε0K

=q(q2AKε0)=q22AKε0 (where σ2ε0K is the electric field produced by one plate at the location of other).



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