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(B) The electric field $E$ between the plates of a parallel plate capacitor is given by $E = \frac{\sigma}{\epsilon_0}$,where $\sigma$ is the surface

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$F/2$

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(B) The electric field $E$ between the plates of a parallel plate capacitor is given by $E = \frac{\sigma}{\epsilon_0}$,where $\sigma$ is the surface charge density.This field is the superposition of the fields produced by the two individual plates,each contributing $E_{plate} = \frac{\sigma}{2\epsilon_0}$.Since $F = qE$,the force $F$ experienced by a test charge $q$ is $F = q \left( \frac{\sigma}{\epsilon_0} \right)$.When one plate is removed,the electric field becomes $E' = \frac{\sigma}{2\epsilon_0}$,which is half of the original field.Therefore,the new force $F'$ experienced by the test charge is $F' = qE' = q \left( \frac{\sigma}{2\epsilon_0} \right) = \frac{F}{2}$.

When a test charge placed in the middle of a parallel plate capacitor experiences a force $F$,what will be the force experienced by the same test charge if one plate is removed?

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