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  • At a point A, there is an electric field of 500V/m and potential difference of 300V. The distance between the point charge and A is -Option: 1 6 m</p

At a point A, there is an electric field of 500V/m and potential difference of 300V. The distance between the point charge and A is -

Option: 1

6 m


Option: 2

36 m


Option: 3

12 m


Option: 4

144 m


Answers (1)

best_answer

Electric field , E = \frac{kq}{r^{2}}

Electric potential ,  V = \frac{kq}{r}

Now, \frac{V}{E} = \frac{kq}{r} \times \frac{r^{2}}{kq} = r

Now, r = \frac{V}{E} = \frac{3000}{500} = 6 Volt

Posted by

avinash.dongre

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