\(C = 4\pi {\varepsilon _0}R = \frac{{0.5 \times {{10}^{ - 3}}}}{{9 \times {{10}^9}}} = \frac{1}{{18}} \times {10^{ - 12}}\,F\)
સપાટી પરથી બહાર નીકળતાં વિદ્યુતભારોનો દર
\( = \frac{{80}}{{100}} \times 6.25 \times {10^{10}} \times 1.6 \times {10^{ - 19}} = 8 \times {10^{ 9}}\ C/s\)
તેથી \(q = (8 \times 10^{-9})t\) અને \(q = CV\)
\(8 \times {10^{ - 9}} \times t = \frac{1}{{18}} \times {10^{ - 12}} \times 1\,\,\, \Rightarrow \,t = \frac{{{{10}^{ - 12}}}}{{8 \times {{10}^{ - 9}} \times 18}} = \frac{{{{10}^{ - 3}}}}{{144}} = 6.95\,\mu s\)
$\varepsilon(x)=\varepsilon_{0}+k x, \text { for }\left(0\,<\,x \leq \frac{d}{2}\right)$
$\varepsilon(x)=\varepsilon_{0}+k(d-x)$, for $\left(\frac{d}{2} \leq x \leq d\right)$