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

\frac{2 C}{x}

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\frac{C}{x}

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\frac{2 C}{x^{2}}

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\frac{C}{2 x^{2}}

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Solution

The correct option is D $\frac{C}{x}$
$V = - \int_{\infty}^{x} E d x = - \int_{\infty}^{x} \frac{C}{x^{2}} d x = \frac{C}{x}$

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The gravitational field due to a mass distribution is E=Cx2 in x-direction. Here C is constant. Taking the gravitational potential to be zero at infinity, potential at x is:

The correct option is D CxV=−∫x∞Edx=−∫x∞Cx2dx=Cx

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

The correct option is D $\frac{C}{x}$
$V = - \int_{\infty}^{x} E d x = - \int_{\infty}^{x} \frac{C}{x^{2}} d x = \frac{C}{x}$