Question 132 of 180

A force of $10N$ acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be.

A$5N$B$10N$C$20N$DZero

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Answer: (A) $5N$

Electric field due to each plate, ${E}_{1}={E}_{2}=\dfrac{\sigma }{2{\epsilon }_{0}}=\dfrac{Q}{2A{\epsilon }_{0}}$
Net electric field between the plates, ${E}_{net}={E}_{1}+{E}_{2}=\dfrac{Q}{A{\epsilon }_{0}}$
Force on charged particle between the plates, ${F}_{1}=q{E}_{net}=\dfrac{qQ}{A{\epsilon }_{0}}=10N$

Now, in second case, the net electric field, $E=\dfrac{\sigma }{2{\epsilon }_{0}}=\dfrac{Q}{2A{\epsilon }_{0}}=5N$
Force on charged particle, ${F}_{2}=qE=\dfrac{qQ}{2A{\epsilon }_{0}}=5N$

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