Capacitors
7.0 Mechanical force on the charged conductor
7.0 Mechanical force on the charged conductor
We know that similar charges repel each other, hence the charge on any part of surface of the conductor is repelled by the charge on its remaining part. The surface of the conductor thus experiences a mechanical force. The electric field at any point $P$ near the conductor surface can be assumed as due to a small part of the surface of area say $\Delta s$ immediately in the neighborhood of the point under consideration and due to the rest of the surface.
Let $\overrightarrow {{E_1}} $ and $ \overrightarrow {{E_2}} $ be the field intensities due to these parts respectively. Then, total electric field, $$\overrightarrow E = \overrightarrow {{E_1}} + \overrightarrow {{E_2}} $$$\vec E$ has a magnitude $\frac{\sigma }{{{\varepsilon _ \circ }}}$ at any point $P$ just outside the conductor and is zero at point $Q$ just inside the conductor. Thus, at $P$,
$${E_1} + {E_2} = \frac{\sigma }{{{\varepsilon _ \circ }}}$$
and at $Q$, $${E_1} - {E_2} = 0$$
Therefore $${E_1} = {E_2} = \frac{\sigma }{{2{\varepsilon _{_o}}}}$$
Hence, the force experienced by small surface of are $\Delta S$ due to the charge on the rest of the surface $$F = q{E_2} = \frac{{({\sigma ^2})(\Delta S)}}{{2{\varepsilon _ \circ }}}$$ $$\frac{{Force}}{{Area}} = \frac{F}{{\Delta S}} = \frac{{{\sigma ^2}}}{{2{\varepsilon _ \circ }}} = \frac{1}{2}{\varepsilon _ \circ }{E^2}$$ $$\frac{{Force}}{{Area}} = \frac{1}{2}{\varepsilon _ \circ }{E^2}$$
