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Question

The gravitational potential due to a mass distribution is V=Ax2+a2. The gravitational field due to the mass distribution is

A
Ax(x2+a2)32
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B
2Ax(x2+a2)32
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C
4Ax(x2+a2)32
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D
8Ax(x2+a2)32
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Solution

The correct option is A Ax(x2+a2)32
Gravitation field is given as,
E=Ex^i+Ey^j+Ez^k

Here, we will consider only Ex^i, as the potential does not have any y and z components. Hence Ey=0 and Ez=0.

So, Ex=dVdx
where, V=A(x2+a2)1/2 (given)
Ex=dVdx=d[A(x2+a2)1/2]dxEx=A(12)(x2+a2)3/2.2x
Ex=Ax(x2+a2)3/2

Hence, option A is the correct option.

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