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.