If Q is the charge on the plates of a capacit | vedclass

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(D) The electric field produced by one plate at the location of the other plate is $E = \frac{\sigma}{2 \varepsilon_0}$,where $\sigma = \frac{Q}{A}$ i

Vedclass · Accepted Answer

Both $A$ and $B$

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(D) The electric field produced by one plate at the location of the other plate is $E = \frac{\sigma}{2 \varepsilon_0}$,where $\sigma = \frac{Q}{A}$ is the surface charge density.The force of attraction $F$ on a charge $Q$ due to this electric field is $F = Q E = Q \left( \frac{Q}{2 \varepsilon_0 A} \right) = \frac{Q^2}{2 \varepsilon_0 A}$.Since the capacitance of a parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$,we can write $\varepsilon_0 A = C d$.Substituting this into the force expression: $F = \frac{Q^2}{2 C d}$.Using the relation $Q = CV$,we get $F = \frac{(CV)^2}{2 C d} = \frac{C^2 V^2}{2 C d} = \frac{1}{2} \frac{C V^2}{d}$.Thus,both expressions $\frac{Q^2}{2 \varepsilon_0 A}$ and $\frac{C V^2}{2 d}$ are correct. Therefore,the correct option is $D$.

If $Q$ is the charge on the plates of a capacitor of capacitance $C$,$V$ is the potential difference between the plates,$A$ is the area of each plate,and $d$ is the distance between the plates,then the force of attraction between the plates is:

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