Two 3.0 cm X 3.0 cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 V battery.

(a)What is the charge on each electrode? (pC)

(b)What is the potential difference between electrodes? (V)

(c)While the plates are still connected to the battery, inselated handles are used to pull them apart to a new spaceing of 2.0mm. What are the charge on each electrode? (pC)

(d) While the plates are still connected to the battery, inselated handles are used to pull them apart to a new spaceing of 2.0mm. What is the potential difference between electrodes? (V)

a) 7.1685 * 10⁻¹¹C

b) 9.0V

c) 3.58425 * 10⁻¹¹C

d) 9.0 V

Explanation:

a) the charge on the electrode is given by the formula

Q = CV where

C = Capacitance

and V is the Potential Difference

but

C = e₀A/d,

e₀ = dielectric constant, 8.85 × 10⁻¹² F/m

A = area of the plates, (A = L X B), in meters, 0.03m × 0.03m = 0.0009m²

d = distance between the plates, 1.0mm or 0.001m

hence

C = 7.965 * 10⁻¹² F

C = (7.965 * 10⁻¹² × 10¹²)pF        [1F = 10¹²pF]

C = 7.965pF

Now,

Q = CV

Q = 7.965 * 10⁻¹² × 9.0

Q = 7.1685 * 10⁻¹¹C

(b)

Potential Difference, = 9.0V

(c)

New d = 2.0mm or 2 * 10⁻³m

new capacitance, C = e₀A/d

C = ( 8.85 * 10⁻¹² × 9 * 10⁻⁴)/2 * 10⁻³

C = 3.9825 * 10⁻¹²F

C = 3.98pF

Q = CV

Q = 3.9825 * 10⁻¹² × 9.0

Q = 3.58425 * 10⁻¹¹C

Q = 3.58 * 10⁻¹¹C

(d)

the potential difference between the electrode still remains 9.0 V because it is from the battery

answer:

make you upset

explanation: