Electric potential at a point ‘P’ due to a point charge of 1×10−9C is 9 volt, the distance of ‘P’ from the point charge is ______ cm (Assume14π∈o=9×109Nm2C2)

Potential due to a point charge at a distance ‘r’ is V=9×109qr⇒9=9×109×1×10−9r ⇒r=1m⇒r=100cm

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Electric potential at a point ‘P’ due to a point charge of $1 \times 10^{- 9} C$ is 9 volt, the distance of ‘P’ from the point charge is ______ cm $(A s s u m e \frac{1}{4 π \in_{o}} = 9 \times 10^{9} \frac{N m^{2}}{C^{2}})$

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Potential due to a point charge at a distance ‘r’ is

$V = 9 \times 10^{9} \frac{q}{r} \Rightarrow 9 = 9 \times 10^{9} \times \frac{1 \times 10^{- 9}}{r} \Rightarrow r = 1 m \Rightarrow r = 100 c m$

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