Question

A charge of \[4 \times 10^{-7}\ \text{C}\] is located 9 cm away from a point P. Taking \[\frac{1}{4\pi\varepsilon_0}=9 \times 10^9\ \text{N m}^2\text{C}^{-2}\], the electric potential at P is:

पर्याय

• \[4 \times 10^4\ \text{V}\]

• \[4 \times 10^3\ \text{V}\]

• \[9 \times 10^4\ \text{V}\]

• \[4 \times 10^5\ \text{V}\]

Answer 1

Using \[V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}=9\times10^9\times\frac{4\times10^{-7}}{0.09}\], we get \[V = 4 \times 10^4\ \text{V}\]. Here \[r = 9\ \text{cm} = 0.09\ \text{m}\] must be converted to SI units before substitution.