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16 September, 09:14

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

What is the magnitude of the potential difference between the plates?

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  1. 16 September, 09:26
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    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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