On the $$x-axis$$ and at a distance x from the origin, the gravitational field due to a mass distribution is given by $$Axx2+a232$$ 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 :

Question

On the $$x-axis$$ and at a distance x from the origin, the gravitational field due to a mass distribution is given by  $$Axx2+a232$$ 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 :

Infinity Learn · Accepted Answer

Given $$Ex=Axx2+a23/2$$∫$$v=$$∞$$v=vxdv=-$$∫∞xEdx∵$$E=dvdrV(x)-V($$∞$$)=-$$∫∞$$xAxx2+a23/2dx$$   $$----(1)Let$$ $$x2+a2=zdifferentiate$$ bothsides with respective x we $$get2xdx+0=dz$$⇒$$xdx=dz2----(2)From$$ (1) and (2) Given $$V($$∞$$)=0$$⇒$$V(x)-0=-A2$$∫∞$$xdzz3/2=-A2$$∫∞$$xz-32dz=-A2z-3/2+1-3/2+1=-A2x2+a2-1/2-12$$∞$$x=Ax2+a2-1/2$$

On the x-axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by $\frac{A x}{{(x^{2} + a^{2})}^{\frac{3}{2}}}$ 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 :

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On the x-axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by Axx2+a232 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 :

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On the x-axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by $\frac{A x}{{(x^{2} + a^{2})}^{\frac{3}{2}}}$ 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 :