Question

Electric potential at 0.5m from a point
charge is 50v. What is the magnitude and
sign of charge​

Answer 1

Point charge will be equal to $2.777\times 10^{-9}C$

Explanation:

We have given that electric potential; at a point 0.5 m form the point charge is 50 volt

So potential v = 50 volt

Distance from point charge r = 0.5 m

Electric potential due to a point charge is equal to $V=\frac{1}{4\pi \epsilon _0}\frac{Q}{r}$, here Q is point charge and r is distance of the point where we have to find electric potential from the charge

So $V=\frac{KQ}{r}$

$50=\frac{9\times 10^9\times\ Q}{0.5}$

$Q=2.777\times 10^{-9}C$

So point charge will be equal to $2.777\times 10^{-9}C$

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