The plate area is doubled, and the plate separation is reduced to half its initial separation. What is the new charge on the neg
1 answer:
Answer:
Q = 4 Q₀
Explanation:
This is an exercise on capacitors, where the capacitance is
C =
if we apply the given conditions
C = \epsilon_{o} \ \frac{2A}{0.5d}
C = 4 \epsilon_{o} \ \frac{A}{d}
let's call the capacitance Co with the initial values
C₀ = \epsilon_{o} \ \frac{A}{d}
C = 4 C₀
The charge on each plate of a capacitor is
Q = C ΔV
If the potential difference is maintained, the new charge is
Q = 4 C₀ ΔV
let's call
Q₀ = C₀ ΔV
we substitute
Q = 4 Q₀
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