A force of 10 ; N acts on a charged particle | vedclass

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Question

(A) The electric field between the plates of a parallel plate capacitor is given by $E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A \epsilon_0}$.The force

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The electric field between the plates of a parallel plate capacitor is given by $E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A \epsilon_0}$.The force on a charged particle $q$ is $F = qE = \frac{qQ}{A \epsilon_0} = 10 \; N$.When one plate is removed,the remaining plate acts as a single charged sheet. The electric field due to a single charged sheet is $E^{\prime} = \frac{\sigma}{2 \epsilon_0} = \frac{Q}{2 A \epsilon_0}$.Therefore,the new force acting on the particle is $F^{\prime} = qE^{\prime} = \frac{qQ}{2 A \epsilon_0} = \frac{1}{2} \left( \frac{qQ}{A \epsilon_0} ight) = \frac{1}{2} 	imes 10 \; N = 5 \; N$.

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

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