Physics, asked by linju595, 1 month ago

A parallel plate air capacitor has a capacitor of 245pF has a charge of magnitude 0.148uc on each plate. Find the potential difference and electric field intensity between the plates if the distance between the plates is 5 mm

Answers

Answered by jawedkha606

1

Answer:

We know that for capacitor, Q=CV

The potential between plates is V=

C

Q

=

245×10

−12

0.148×10

−6

=604V

Explanation:

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