Question

Two 2.0 cm by 2.0 cm metal electrodes are spaced 1.0 mm apart and connected by wires of the terminals of a 9.0 V battery.

a. What are the charge on each electrode and the potential difference between them?
b. If the wires are disconnected, and insulating handles are used to pull the plates apart to a new spacing of 2.0 mm, what are the charge on each electrode and the potential difference between them?
c. If instead, the plates from part a. remained connected to the battery while the insulating handles pull them apart to their new spacing of 2.0 mm, what are the charge on each electrode and the potential difference between them?

Answer 1

Answer:

Explanation:

Area of electrodes, A = 2 cm x 2 cm = 4 cm²

Separation between electrodes, d = 1 mm

Voltage, V = 9 V

(a)

Let C is the capacitance between the electrodes

$C = \frac{\epsilon _{0}A}{d}$

$C = \frac{8.854\times 10^{-12}\times 4\times 10^{-4}}{1\times 10^{-3}}$

C = 3.54 x 10^-12 F

Let q be the charge on each of the electrode

q = C x V

q = 3.54 x 10^-12 x 9 = 3.2 x 10^-11 C

(b)

As, the battery is disconnected the charge on the electrodes remains same.

(c)

As the battery is connected the voltage is same.

capacitance is change.

As the distance is doubled, the capacitance becomes half and the charge is also halved. q' = q/2 = 1.6 x 10^-11 C