# solved problems on electrostatics

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1) The plate area of parallel plate capacitor is 0.01 $m^2$. The distance between the plate is 2.5 cm. The insulating medium is air. Find capacitance, if the space between the plates is filled with an insulating material of relative permittivity 5?

$C = \dfrac{\epsilon_0 \epsilon_r A }{d}$

$C = \dfrac{8.854 \times 10^{-12} \times 1 \times 0.01}{2.5 \times 10^{-1}}$

C = $3.5416 \times 10^{-12}$ F

for $\epsilon_r = 5$
$C = \dfrac{\epsilon_0 \epsilon_r A }{d}$

$C = \dfrac{8.854 \times 10^{-12} \times 5 \times 0.01}{2.5 \times 10^{-1}}$

$C = 1.7708 \times 10^{-11}$

2) A parallel plate capacitor has a plate area of 500 $cm^2$, a plate separation of 0.15 mm and air as a dielectric. Find the capacitance. Also find the voltage between the plates the electric flux density and electric field strength, if the capacitor has a charge of 2 $\mu$ C.

$C = \dfrac{\epsilon_0 \epsilon_r A }{d}$

$C = \dfrac{8.854 \times 10^{-12} \times 1 \times 500 \times 10^{-4}}{0.15 \times 10^{-3}}$

$C = 2.9513 \times 10^{-9} F$

$Q = CV$

$V = \dfrac{Q}{C}$

$V = \dfrac{2 \times 10^{-6}}{2.9513 \times 10^{-9} }$

$V = 677.6674$ V

$D = \dfrac{2 \times 10^{-6}}{500 \times 10^{-4}}$

$D = 4 \times 10^{-5} \; C/m^2$

$E = \dfrac{V}{d}$

$E = \dfrac{677.6674}{0.15 \times 10^{-3}}$

$E = 4517782.667 V/m$

Q) A 5 $\mu$F capacitor is charged to a potential difference of 110 V and then connected in parallel with an uncharged 3 $\mu$F capacitor. Calculate the potential difference across the parallel capacitor.

$Q_1 = C_1 \times V_1 = 5 \times 10^{-6} \times 110 = 5.5 \times 10^{-4} C$

Both the capacitors are connected in parallel. Therefore voltage across both the capacitor will remain same.

$V_1 = V_2$

$\dfrac{q_1}{C_1} = \dfrac{q_2}{C_2}$

Now capacitor $C_2$ is uncharged, therefore the charge will flow from $C_1$ to $C_2$ and this flow of charge will stop when the potential difference across $C_1$ and $C_2$ will become equal i.e. $V_1 = V_2$. Capacitor $C_1$ will loose charge say $q_2$ and capacitor $C_2$ will gain that charge $q_2$. Hence new charge on $C_1$ will become $q_1$.

$\therefore$ Charge on $C_1$ will be $q_1 = Q_1 – q_2$

$\dfrac{Q_1 – q_2}{C_1} = \dfrac{q_2}{C_2}$

$\dfrac{(5.5 \times 10^{-4}) – q_2}{5 \times 10^-6} = \dfrac{q_2}{3 \times 10^-6}$

$q_2 = 2.06525 \times 10^{-4}$

$V_1 = V_2 = \dfrac{q_2}{C_2} = 68.75 V$