The charge on a parallel plate capacitor of c | vedclass

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Question

(B) The electric field intensity $E$ between the plates of a parallel plate capacitor is given by the formula:$E = \frac{\sigma}{\varepsilon_0} = \fra

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$\frac{Q}{Ct}$

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(B) The electric field intensity $E$ between the plates of a parallel plate capacitor is given by the formula:$E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A \varepsilon_0}$  --- $(i)$The capacitance $C$ of a parallel plate capacitor with plate area $A$ and separation distance $t$ is given by:$C = \frac{A \varepsilon_0}{t}$From this,we can write $A \varepsilon_0 = Ct$ --- $(ii)$Substituting the value of $A \varepsilon_0$ from equation $(ii)$ into equation $(i)$:$E = \frac{Q}{Ct}$Therefore,the electric field intensity is $\frac{Q}{Ct}$.

The charge on a parallel plate capacitor of capacity $C$ is $Q$. The electric field intensity between its two plates,which are separated by a distance $t$,is:

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