The force acting on a charged particle placed | vedclass

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Question

(B) Initially,the electric field between the plates of a parallel plate capacitor is $E = \frac{\sigma}{\varepsilon_0}$,where $\sigma$ is the surface

Vedclass · Accepted Answer

$F/2$

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(B) Initially,the electric field between the plates of a parallel plate capacitor is $E = \frac{\sigma}{\varepsilon_0}$,where $\sigma$ is the surface charge density. The force on the particle is $F = qE = \frac{q\sigma}{\varepsilon_0}$.When one plate is removed,the capacitor effectively becomes a single charged sheet. The electric field due to a single charged sheet is given by $E' = \frac{\sigma}{2\varepsilon_0}$.Therefore,the new force acting on the particle is $F' = qE' = q \left( \frac{\sigma}{2\varepsilon_0} \right) = \frac{1}{2} \left( \frac{q\sigma}{\varepsilon_0} \right) = \frac{F}{2}$.

The force acting on a charged particle placed between the plates of a charged parallel plate capacitor is $F$. If one plate of the capacitor is removed,then the force acting on the same particle will become:

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