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A 5 C charge experiences a force of 2000 N when moved between two points separated by a distance of 2 cm in a uniform electric field. The potential difference between the two points is

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8 V

200 V

800 V

20000 V

answer is A.

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Detailed Solution

$E = \frac{F}{q_{0}} = \frac{2000 N}{5 C} = 400 \frac{N}{C}$

In a uniform field,

$Δ V = E . d = (400 \frac{N}{C}) (2 \times 10^{- 2} m)$

$= 8 V$

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