The gravitational field due to a mass distribution is E = $\frac{A}{x^{2}}$ in x-direction. Here, A is a constant. Taking the gravitational potential to be zero at infinity, potential at x is

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Solution

The correct option is C

$V (x) = - \int_{\infty}^{x} E d x = - \int_{\infty}^{x} \frac{A}{x^{2}} d x = \frac{A}{x}$

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The gravitational field due to a mass distribution is E = Ax2 in x-direction. Here, A is a constant. Taking the gravitational potential to be zero at infinity, potential at x is

The correct option is C V(x)=−∫x∞Edx=−∫x∞Ax2dx=Ax

The gravitational field due to a mass distribution is E = $\frac{A}{x^{2}}$ in x-direction. Here, A is a constant. Taking the gravitational potential to be zero at infinity, potential at x is

The correct option is C

$V (x) = - \int_{\infty}^{x} E d x = - \int_{\infty}^{x} \frac{A}{x^{2}} d x = \frac{A}{x}$