Question

A dielectric slab of constant k = 3 is inserted inside gap of parallel plate capacitor of plate area $L\times L$ and separation of d at speed u. Power dissipation across resistance R is

• A. $\frac{{\epsilon }_0^2{L}^2{V}^2{u}^{2}R}{d^2}$
• B. $\frac{2{\epsilon }_0^2{L}^2{V}^2{u}^{2}R}{d^2}$
• C. $\frac{3{\epsilon }_0^2{L}^3{V}^2{u}^{2}R}{d^2}$
• D. $\frac{4{\epsilon }_0^2{L}^3{V}^2{u}^{2}R}{d^2}$

Answer 1

The correct option is D $\frac{4{\epsilon }_0^2{L}^3{V}^2{u}^{2}R}{d^2}$


$Leq=C_1+C_2=\frac{{\epsilon }_{0}L\mu K}{d}+\frac{}{}\frac{{\epsilon }_{0}L(L-\mu }{d}$
$q=leqV=\frac{{\epsilon }_{0}LV}{d}[L+(k-1\left)x\right]$
So $i=\frac{dq}{dt}=\frac{{\epsilon }_{0}LV}{d}(k-1)\frac{dx}{dt}=\frac{2{\epsilon }_{0}LVu}{d}$
Power = $i^{2}R=\frac{4{\epsilon }_0^2{L}^3{V}^2{u}^{2}R}{d^2}$