Question

The gravitational potential due to a mass distribution is $V=\frac{A}{\sqrt{x^2+a^2}}$. The gravitational field due to the mass distribution is

• A. $\frac{Ax}{(x^2+a^2{)}^{\frac{3}{2}}}$
• B. $\frac{2Ax}{(x^2+a^2{)}^{\frac{3}{2}}}$
• C. $\frac{4Ax}{(x^2+a^2{)}^{\frac{3}{2}}}$
• D. $\frac{8Ax}{(x^2+a^2{)}^{\frac{3}{2}}}$

Answer 1

The correct option is A $\frac{Ax}{(x^2+a^2{)}^{\frac{3}{2}}}$

Gravitation field is given as,
$\overrightarrow{E}=E_x\hat{i}+E_y\hat{j}+E_z\hat{k}$

Here, we will consider only $E_x\hat{i}$, as the potential does not have any $y$ and $z$ components. Hence $E_y=0$ and $E_z=0$.

So, $E_x=-\frac{dV}{dx}$
where, $V=A(x^2+a^2{)}^{-1/2}$ (given)
$E_x=-\frac{dV}{dx}=-\frac{d\left[A\right(x^2+a^2{)}^{-1/2}]}{dx}{E}_x=-A(-\frac{1}{2})(x^2+a^2{)}^{-3/2}.2x$
$E_x=\frac{Ax}{(x^2+a^2{)}^{3/2}}$

Hence, option A is the correct option.