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A potential difference of 3000 V is maintained between two large parallel plates 6cm apart. The force on a proton midway between the plates is

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8 × 10^–15 N

8 × 10^–13 N

8 × 10^–6 N

8 × 10^–12 N

answer is B.

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Detailed Solution

$F = Eq = \frac{V}{d} q; F = \frac{3000}{6 \times 10^{- 2}} \times 1 .6 \times 10^{- 19}$
= F = 8 × 10^–15 N.

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