Question

A parallel plate capacitor of capacitance Co has plates of area A with separation d between them. When it is connected to a battery of voltage Vo, it has charge of magnitude Qo on its plates. It is then disconnected from the battery and the plates are pulled apart to a separation 2d without discharging them. After the plates are 2d apart, the new capacitance and the potential difference between the plates are: (Show all steps) [2 marks] a. ½ Co, ½ Vo b. ½ Co, 2Vo c. Co, Vo d. Co, 2Vo e. 2Co, 2Vo

Answer 1

Answer:

b. 1/2·C₀, 2·V₀

Explanation:

The capacitance on the parallel plate capacitor = C₀

The area of the plates  = A

The voltage on the battery = V₀

The magnitude of the charge on the plate = Q₀

The new distance between the plates = 2·d

From an online physics source, we have;

$C_0 = \epsilon_0 \times \dfrac{A}{d}$

Where;

ε₀ = Constant

A = The area of the plates

With the new distance, 2·d, we get;

$C_{new} = \epsilon_0 \times \dfrac{A}{2\cdot d} = \dfrac{1}{2} \times \epsilon_0 \times \dfrac{A}{d} = \dfrac{1}{2} \times C_0$

Therefore;

$The \ new \ capacitance \ C_{new} = \dfrac{1}{2} \times C_0$

The potential difference, 'V', is given as follows;

$C = \dfrac{Q}{V}$

Therefore;

$V = \dfrac{Q}{C}$

Given that Q = Q₀, we get;

$V = \dfrac{Q_0}{\dfrac{1}{2} \times C_0} = 2 \times \dfrac{Q_0}{C_0} = 2 \times V_0$

∴ V = 2 × V₀

The new potential difference, V = 2·V₀

Therefore, after the plates are 2·d apart, the new capacitance and potential difference between the plates are;

1/2·C₀, 2·V₀.