Phy 315- PhysicsI 1. (a) Write down the equation for the capacitance of a parall

Question

Phy 315- PhysicsI 1. (a) Write down the equation for the capacitance of a parallel plate capacitor with dielectric material inside. Identify the meaning of each symbol in the equation. (b) A parallel plate capacitor has plates with dimensions 1.00 cm × 40.0 cm. The separation between plates is 0.200 m. If the capacitor filled with Mica with dielectric constant of 7.00 calculate the capacitance. (c) If the plates were connected to 12.0 V battery, calculate the magnitude of the charge on a plate. (d) Calculate the total energy stored in the capacitor when connected to 12.0 V battery. (e) If you increase the voltage between the capacitor what would happen to the following quantities. Capacitance- Reason: Energy stored- Reason: 1 1 Page

Explanation / Answer

(a) Capacitance, C = (k * epcilon0 * A) / d

where, k = relative permittivity of the dielectric material between the plates

epcilon0 = 8.854 x 10^-12 F/m

A = Area of the plate

d = separation of the plate

(b) Here, A = 1.0 x 40.0 = 40.0 x 10^-4 m^2

d = 0.20 m

k = 7.0

So, C = (7.0 x 8.854 x 10^-12 x 40.0 x 10^-4) / 0.20 = 1.24 x 10^-12 F

(c) Charge, Q = C*V = 1.24 x 10^-12 x 12 = 14.88 x 10^-12 C = 1.488 x 10^-11 C

(d) Total Energy stored -

E = (1/2)*C*V^2 = 0.5 x 1.24 x 10^-12 x 12^2 = 89.28 x 10^-12 J = 8.928 x 10^-11 J

(e) If we increase the voltage between the capacitor then -

There will be NO EFFECT on the capacitance.

Reason, Capacitance C = (k * epcilon0 * A) / d

the expression is independent of V.

Energy stored E WILL INCREASE.

Reason -

E = (1/2)*C*V^2

when V is increase then E will also increase.   

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