Question

On the $x$ -axis and a dsitance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{ Ax }{\left( x ^{2}+ a ^{2}\right)^{3 / 2}}$ in the x-direction. The magnitude of gravitational potential on the x-axis at a distance $x$, taking its value to be zero at infinity, is :

• A. $\frac{A}{\left(x^{2}+a^{2}\right)^{1 / 2}}$
• B. $\frac{A}{\left(x^{2}+a^{2}\right)^{3 / 2}}$
• C. $A \left( x ^{2}+ a ^{2}\right)^{3 / 2}$
• D. $A \left( x ^{2}+ a ^{2}\right)^{1 / 2}$

Answer 1

Answer: (A)

Given $E _{ G }=\frac{ Ax }{\left( x ^{2}+ a ^{2}\right)^{3 / 2}}, V _{\infty}=0$
$\int\limits_{ V _{\infty}}^{ v _{ x }} d V =-\int\limits_{\infty}^{ x } \vec{ E }_{ G } \cdot \vec{ d }_{ x }$
$V _{ x }- V _{\infty}=-\int\limits_{\infty}^{ x } \frac{ Ax }{\left( x ^{2}+ a ^{2}\right)^{3 / 2}} d x$
put $x ^{2}+ a ^{2}= z$
$2 x dx = d z$
$V _{ x }-0=-\int \limits_{\infty}^{ x } \frac{ Adz }{2( z )^{3 / 2}}=\left[\frac{ A }{ z ^{1 / 2}}\right]_{\infty}^{ x }=\left[\frac{ A }{\left( x ^{2}+ a ^{2}\right)^{1 / 2}}\right]_{\infty}^{ x }$
$V _{ x }=\frac{ A }{\left( x ^{2}+ a ^{2}\right)^{1 / 2}}-0=\frac{ A }{\left( x ^{2}+ a ^{2}\right)^{1 / 2}}$