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.

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