A uniform electric field of $10 N C^{- 1}$ exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a height of 50 cm.

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Solution

$E = 10 N / C, s = 50 c m = \frac{1}{2}$
or $v = e . s = 10 \times \frac{1}{2}$

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A uniform electric field of 10NC−1 exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a height of 50 cm.

E=10N/C,s=50cm=12 or v=e.s=10×12

A uniform electric field of $10 N C^{- 1}$ exists in the vertically downward direction. Find the increase in the electric potential as one goes up through a height of 50 cm.

$E = 10 N / C, s = 50 c m = \frac{1}{2}$
or $v = e . s = 10 \times \frac{1}{2}$