The potential at a point 0.1 m from an isolated point charge is +100 Volt. The nature of the point charge is-

a) positive

b) negative

c) zero

d) either positive or zzero

and why?????

plzzz tell..

The correct answer is : (a) Positive

the relation between potential $V$, charge $q$ and the distance $r$ is given as,
 $V=\frac{Kq}{r}$
or,
 $V=\frac{1}{4\pi {\epsilon }_o}\frac{q}{r}$
where, ${\epsilon }_o$ is an electric constant and the value of Coulombs' constant is $\frac{1}{4\pi {\epsilon }_o}=9\times {10}^9 N{m}^2/C^2$
Now rearranging the equation for charge we have,
 $q=\left(4\pi {\epsilon }_o\right)\times V\times r$
On substituting  the values we get,
 $q=\left(9\times {10}^9 N{m}^2/C^2\right)\times 100 V\times 0.1 m = + 9 X {10}^{10} C$
Thus, the magnitude of charge is positive $q=+9\times 10$¹⁰ C.