JEE Main & Advanced Physics Electrostatics & Capacitance Force on a Charged Conductor

To find force on a charged conductor (due to repulsion of like charges) imagine a small part XY to be cut and just separated from the rest of the conductor MLN. The field in the cavity due to the rest of the conductor is \[{{E}_{2}},\] while field due to small part is \[{{E}_{1}}\]. Then

Inside the conductor \[E={{E}_{1}}-{{E}_{2}}=0\] or \[{{E}_{1}}={{E}_{2}}\]

Outside the conductor \[E={{E}_{1}}+{{E}_{2}}=\frac{\sigma }{{{\varepsilon }_{0}}}\]

Thus \[{{E}_{1}}={{E}_{2}}=\frac{\sigma }{2{{\varepsilon }_{0}}}\]

(1) To find force, imagine charged part XY (having charge \[\sigma \,dA\] placed in the cavity MN having field \[{{E}_{2}}\]). Thus force \[dF=(\sigma \,dA){{E}_{2}}\] or \[dF=\frac{{{\sigma }^{2}}}{2{{\varepsilon }_{0}}}dA\]. The force per unit area or electrostatic pressure \[p=\frac{dF}{dA}=\frac{{{\sigma }^{2}}}{2{{\varepsilon }_{0}}}\]

(2) The force is always outwards as \[{{(\pm \sigma )}^{2}}\] is positive i.e., whether charged positively or negatively, this force will try to expand the charged body. [A soap bubble or rubber balloon expands on charging to it (charge of any kind + or \[-\])].