A parallel plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148 μC on each plate. The plates are 0.328 mm apart then the potential difference between the plates would beChoose answer: 604 V 620 V 600 V 588 V
Question
A parallel plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148 μC on each plate. The plates are 0.328 mm apart then the potential difference between the plates would beChoose answer: 604 V 620 V 600 V 588 V
Solution
The potential difference (V) across a capacitor can be calculated using the formula:
V = Q / C
where: Q is the charge on the capacitor, and C is the capacitance of the capacitor.
Given: Q = 0.148 μC = 0.148 x 10^-6 C (since 1 μC = 10^-6 C) C = 245 pF = 245 x 10^-12 F (since 1 pF = 10^-12 F)
Substituting these values into the formula, we get:
V = (0.148 x 10^-6) / (245 x 10^-12) V = 604 V
So, the potential difference between the plates is 604 V.
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