The potential difference between a pair of oppositely charged parallel plates is 398 V. If the spacing between the plates is dou bled without altering the charge on the plates, what is the new potential difference between the plates? Answer in units of V.
1 answer:
Answer:
Explanation:
capacitance of parallel plate capacitor
c = ε A / d , d is distance between plates , A is surface area , ε is constant
As d becomes two times , Capacitance c = 1/ 2 times ie c / 2
potential V = Q / C
Q is constant , potential
v = Q / c /2
= 2 . Q / C
= 2 V
So potential difference becomes 2 times.
NEW P D = 398 X 2
= 796 V.
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