Question

1.) A parallel-plate capacitor is constructed from two circular plates with diameter 5.0 cm separated by 0.20 mm. What charge will collect on the plates if the capacitor is connected to a potential difference of 35.0 V?

2.)The capacitor from the previous problem is now disconnected from the voltage source. The separation between the capacitor plates is quadrupled and an insulator is inserted between the plates. The voltage between the plates is measured to be 68.3 V. Find the dielectric constant of the insulator.

3.)A battery with an emf = 6.00 V and an internal resistance of 10.30 ? is connected to a 270 ? resistor. Find the power dissipated in the external resistor?

Answer 1

(1) Diameter d = 5 cm

so, radius r = d/2= 5/2 = 2.5 cm = 0.025 m

so, Cross-sectional area A = pi*r^2 = 3.141*0.025^2 = 0.00196 m^2

The expression for the capacitance C= k*A / d = (1*0.00196) / (0.20*10^-3) = 9.81 F

So, the cahrge on the plates Q = C*V = 9.81*35 = 343.35 C

(2) New capacitance C1 = k1*A / (4*d) = k1*(9.81/4) = k1*2.45

Q will remain constant.

so -

343.35 = k1*2.45 * 68.3

=> k1 = 343.35 / (2.45*68.3) = 2.05

(3) Current I = 6.0 / (10.3 + 270) = 0.0214 A

Power dissipated in the external resistor P = I^2 * 270 = 0.0214^2 * 270 = 0.124 W.