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A parallel-plate capacitor has its plate area doubled and its plate separation reduced by a factor of four at the same time. Using C = epsilon_0 A / d, by what overall factor does the capacitance change?
AIt increases by a factor of eight
BIt increases by a factor of two
CIt increases by a factor of four
DIt increases by a factor of six
Answer & Solution
Correct answer: A. It increases by a factor of eight
1. Capacitance is directly proportional to area, so doubling the area multiplies C by 2.
2. Capacitance is inversely proportional to separation, so cutting the separation by a factor of 4 multiplies C by another factor of 4.
3. Combining both changes multiplies the capacitance by 2 x 4 = 8.
4. Adding the two factors instead of multiplying them, as in option D (2 + 4 = 6), is the usual slip when two proportional changes happen at once.
_Source: OpenStax Physics (CC BY 4.0), Ch 18 "Static Electricity", section 18.5 Capacitors and Dielectrics_
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