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asambeis [7]
4 years ago
9

Two parallel conducting plates are separated by 4.00 cm. The electric field strength between the two plates is 5.70×10⁴ V/m.

Physics
1 answer:
Nookie1986 [14]4 years ago
6 0

Answer:

2280 V

Explanation:

distance between the plates, d = 4 cm = 0.04 m

Electric field strength, E = 5.7 x 10^4 V/m

The formula for potential difference is given by

V = E x d

V = 5.7 x 10^4 x 0.04

V = 2280 V

thus, the potential difference between the plates is 2280 V.

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