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 unchanged. Answer: c) It quadruples.

Step-by-step solution

Write the initial capacitance formula

Write the capacitance of the parallel plate capacitor before any changes as C1, where C1 = εA/d.

Change in dimensions

According to the exercise, the distance between the plates is reduced by half, which means the new distance is d/2. Also, the area of the plates is doubled, so the new area is 2A.

Write the new capacitance formula

Now let's write the formula for the capacitance of the parallel plate capacitor after the changes. We will call this new capacitance C2. So, C2 = ε(2A)/(d/2).

Simplify the new capacitance formula

To find out what happens to the capacitance, we need to simplify C2. Let's do that: C2 = ε(2A)/(d/2) = ε(2A) * (2/d) = ε(4A/d).

Compare the initial and new capacitance

Now let's compare C1 and C2. We know that C1 = εA/d, and we found that C2 = ε(4A/d). To find how C2 relates to C1, we can divide C2 by C1: (C2/C1) = (ε(4A/d)) / (εA/d).

Simplify the division

Simplifying the division expression, we get: (C2/C1) = (ε(4A/d)) / (εA/d) = 4A/d * d/A = 4.

Find the effect on capacitance

As we found that (C2/C1) = 4, it means that the new capacitance C2 is 4 times the initial capacitance C1. Therefore, the capacitance quadruples. Hence, the correct answer is: c) It quadruples.