Question

Plate A of a parallel plate air filled capacitor is connected to a spring having spring constant k
and plate B is fixed. They are held on a frictionless table top as shown in the fig. If a charge +q is
placed on plate A and a charge ?q is placed on plate B, by how much amount does the spring
expand.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Let}{\mathrm{F}}_{\mathrm{s}}=\mathrm{spring}\mathrm{force} \\ {\mathrm{F}}_{\mathrm{e}}=\mathrm{electric}\mathrm{force}\mathrm{that}\mathrm{plate}\mathrm{A}\mathrm{experience}, \\ \mathrm{thus},{\mathrm{F}}_{\mathrm{s}}=-\mathrm{kx}\overrightarrow{\mathrm{i}}.......\left(1\right) \\ \mathrm{and},{\mathrm{F}}_{\mathrm{e}}=\mathrm{qE}\overrightarrow{\mathrm{i}}......\left(2\right) \\ \mathrm{The}\mathrm{electric}\mathrm{force}\mathrm{is}\mathrm{due}\mathrm{to}\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{created}\mathrm{by}\mathrm{plate}\mathrm{B}\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{equation}, \\ \mathrm{E}=\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}=\frac{\mathrm{Q}}{2{\mathrm{a\epsilon }}_0} \\ \mathrm{where}\mathrm{a}\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{one}\mathrm{plate}.\mathrm{Therefore}, \\ \mathrm{kx}=\mathrm{Q}\times \frac{\mathrm{Q}}{2{\mathrm{a\epsilon }}_0} \\ \mathrm{which}\mathrm{gives} \\ \mathrm{x}=\frac{{\mathrm{Q}}^2}{2{\mathrm{a\epsilon }}_0\mathrm{k}} \end{gathered}$$