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Grade: 11
        The electric field and electric potential at a certain point due to a point charge in vacuum are 4000V/m and 8000V respectively . Find the distances of point frome the charge and the magnitude of charge.
one year ago

Answers : (1)

Eshan
askIITians Faculty
2095 Points
							It is given that

E=\dfrac{kq}{r^2}=4000V/m
andV=\dfrac{kq}{r}=8000V

Hence\dfrac{V}{E}=r=2m

\implies q=\dfrac{8000\times 2}{k}=\dfrac{8000\times 2}{9\times 10^9}C=1.77\times 10^{-6}C
one year ago
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