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SOLVED PROBLEMS ON CAPACITOR

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

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

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

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

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

$E = 4517782.667 V/m$

3) Two parallel metal discs, each 100 mm in diameter, are spaced 1 mm apart, the dielectric being air. Calculate the electric field intensity and capacitance.

$A = \dfrac{\pi d^2}{4}$

A = 0.007854 $m^2$

$E = \dfrac{V}{d} = \dfrac{10^3}{10^{-3}}$

$E = 10^6$

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

$C = \dfrac{8.854 \times 10^{-12} \times 1 \times 0.007854}{10^{-3}}$

$C = 6.9539 \times 10^{-11} F$

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

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

$C = \dfrac{8.854 \times 10^{-12} \times 1 \times 0.01}{2.5 \times 10^{-2}}$

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

For $\epsilon_r = 5$

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

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

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

5) 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 $q_1 = Q_1 – q_2$

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

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

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

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

Method 2

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

When the capacitors are connected in parallel, the total capacitance is 8 $\mu F$ and the charge of 0.01 C is divided between the two capacitor.

$Q = CV$

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

$V = \dfrac{5.5 \times 10 ^{-4}}{8 \times 10^{-6}}$

$V = 68.75 V$

6) A 2 $\mu F$ capacitor is charged to 100 V and 3 $\mu F$ capacitor to 200 V.  After disconnecting the supply capacitor terminal of similar polarity are connected together calculate the potential difference between the terminal.

Initial charge on each capacitor.

$Q_{1i} = C_1V_1 = 2 \times 10^{-6} \times 100 = 2 \times 10^{-4} C$

$Q_{2i} = C_2V_2 = 3 \times 10^{-6} \times 100 = 6 \times 10^{-4} C$

Now, $V_1 = V_2$

$\dfrac{Q_{1i} + Q}{C_1} = \dfrac{Q_{2i} – Q}{C_2}$

$\dfrac{2 \times 10^{-4} + Q}{ 2 \times 10^{-6}} = \dfrac{6 \times 10^{-4} – Q}{ 3 \times 10^{-6}}$

$Q = 1.2 \times 10^{-4} C$

$V_1 = V_2 = \dfrac{Q_{1i} + Q}{C_1} = \dfrac{( 2 \times 10^{-4} )+(1.2 \times 10^{-4})}{ 2 \times 10^{-6} } = 160 \;V$

Note:  In the above numerical $Q_{1i} < Q_{2i}$, therefore after connecting similar terminals of the capacitors capacitor $C_2$will lose some charge and $C_1$ will gain some charge

CHARGING AND DISCHARGING OF CAPACITOR

1) A capacitor having capacitance of 4 $\mu F$ is connected in series with a resistance of 1 M$\Omega$ across 200 V DC supply. Find – 1) time constant 2) The initial charging current 3) The time taken by capacitor to raise up to 160 V.

$\tau = RC = 1 \times 10^{6} \times 4 \times 10^{6} = 4 sec$

$I_0 = \dfrac{V}{R} = \dfrac{200}{10^6} = 200 \times 10^{-6} A$

$V_C = V (1 – e^{\frac{-t}{\tau}})$

$160 = 200(1-e^{\frac{-t}{4}})$

$t = 6.4377 sec$