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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 \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

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