Question

the electric potential at 0.1m from a point charge is +50V .what is the magnitude and sign of the charge ?

Answer 1

Answer:

Point charge will be equal to 2.777\times 10^{-9}C2.777×10−9C

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}V=4πϵ01rQ , 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}V=rKQ

50=\frac{9\times 10^9\times\ Q}{0.5}50=0.59×109× Q

Q=2.777\times 10^{-9}CQ=2.777×10−9C

So point charge will be equal to 2.777\times 10^{-9}C2.777×10−9C

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Explanation:

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Answer 2

Answer:

$as \: v = kq \div r \\ q = \: vr \div k \: \: \: \\ q = 0.55nc \\ hope \: it \: would \: help \: u \: and \: plz \\ thank \: my \: answer \: and \: mark \: me \: \\ brainiest........$

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