Question

The parallel-plate air capacitor shown in fig consists of two horizontal conducting plates of equal area A. The bottom plate rests on a fixed support and the top plate is suspended by four springs with spring constant k, positioned at each of the four corners of the plate as shown in Fig. When uncharged, the plates are separated by a distance $z_0$. A battery is connected to the plates and produces a potential difference V between them. This causes the plate separation to decrease to z. Neglect any fringing effects. Find the electrostatic force between the plates ?

• A. $\frac{{\epsilon }_0{A}^2{V}^2}{z^2}$
• B. $\frac{{\epsilon }_0{A}^2{V}^2}{2z^2}$
• C. $\frac{{\epsilon }_0^2{A}^2{V}^2}{z^2}$
• D. $\frac{{\epsilon }_{0}A{V}^2}{2z^2}$

Answer 1

The correct option is D $\frac{{\epsilon }_{0}A{V}^2}{2z^2}$

The parallel plate capacitor is charged by Q due to connecting the battery. Thus $Q=CV$
Where the capacitance $C=\frac{A{ϵ}_0}{z}$. thus, $Q=\frac{A{ϵ}_0}{z}V$
The electrostatic force between the plates, $F=\frac{Q^2}{2A{ϵ}_0}=\frac{(\frac{A{ϵ}_0}{z}V{)}^2}{2A{ϵ}_0}=\frac{A{ϵ}_0{V}^2}{2z^2}$
.