Acharge of magnitude 3×10⁻⁷ C is located at a distance of 0.09 m from a point P. Obtain the work done in bringing a charge of 2×10⁻⁹ C from infinity to the point P.

The correct option is 6×10^-5 J.

To find the work done in bringing a charge from infinity to a point P, we first need to determine the electrostatic potential at point P due to the source charge.

Let the source charge be $q_1$ and the distance from $q_1$ to point P be $r$.
Given data:
Source charge, $q_1=3\times {10}^{-7}\text{ C}$
Distance, $r=0.09\text{ m}$
Charge to be brought from infinity, $q_2=2\times {10}^{-9}\text{ C}$

The electrostatic potential $V$ at point P due to the source charge $q_1$ is given by the formula:
$V=\frac{1}{4\pi {\epsilon }_0}·\frac{q_1}{r}$
where $\frac{1}{4\pi {\epsilon }_0}\approx 9\times {10}^9\text{ N·m}{}^2\text{/C}{}^2$.

Substituting the values into the potential formula:
$V=\frac{(9\times {10}^9)\times (3\times {10}^{-7})}{0.09}$
$V=\frac{27\times {10}^2}{0.09}=3\times {10}^4\text{ V}$

The work done $W$ in bringing the charge $q_2$ from infinity to point P is defined as the product of the charge $q_2$ and the potential $V$ at that point:
$W=q_2·V$

Substituting the values of $q_2$ and $V$:
$W=(2\times {10}^{-9}\text{ C})\times (3\times {10}^4\text{ V})$
$W=6\times {10}^{-5}\text{ J}$

Therefore, the work done in bringing the charge from infinity to point P is $6\times {10}^{-5}\text{ J}$.