Question

i) At what distance the magnitude of electric field and electric potential becomes equal? ​

Answer 1

Answer:

brainly

Explanation:

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Answer 2

At r=1 unit the magnitude of electric field and electric potential becomes equal

Explanation:

Let charge +q is placed at distance r from a point p

Therefore, electric field due to charge q at point p

$E=\frac{kq}{r^2}$

Electric potential at point p

$V=\frac{kq}{r}$

According to question

$V=E$

$\frac{kq}{r}=\frac{kq}{r^2}$

$1=\frac{1}{r}$

$r=1$

At r=1 unit the magnitude of electric field and electric potential becomes equal.

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