Question

The distance between the plates of a parallel plate capacitor is reduced by half and the area of the plates is doubled. What happens to the capacitance? a) It remains unchanged. b) It doubles. c) It quadruples. d) It is reduced by half.

Short answer

a) It halves b) It doubles c) It quadruples d) It remains the same Answer: c) It quadruples

Step-by-step solution

Find the initial capacitance #

To calculate the initial capacitance, we will use the formula C_initial = (ε₀ * A_initial) / d_initial. Here, A_initial and d_initial are the initial area and distance between the plates, respectively. Let's denote the initial capacitance as C₀. C₀ = (ε₀ * A_initial) / d_initial

Find the new capacitance #

Since we're given that the area is doubled and the distance between the plates is reduced by half, the new area and distance will be A_new = 2 * A_initial and d_new = 0.5 * d_initial. Now we can find the new capacitance using the same formula as before: C_new = (ε₀ * A_new) / d_new

Substitute the new values #

Now we need to substitute the new values of area and distance into the equation for the new capacitance: C_new = (ε₀ * (2 * A_initial)) / (0.5 * d_initial)

Simplify the equation #

Now we will simplify the equation to observe any changes in the capacitance: C_new = (ε₀ * 2 * A_initial) / (d_initial * 0.5) C_new = 4 * (ε₀ * A_initial) / d_initial

Compare the new capacitance to the initial capacitance #

We can now compare the new capacitance C_new to the initial capacitance C₀: C_new = 4 * C₀ As we can see, the new capacitance is four times the initial capacitance. Thus, the correct answer is: c) It quadruples.