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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 \muF capacitor is charged to a potential difference of 110 V and then connected in parallel with an uncharged 3 \muF 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

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