Determine The Induced Charge

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Let $\displaystyle Q$ be the induced charge on the sphere.

Now,
Even though this induced charge is non-uniform ,
The potential due to this induced charge at the center of the sphere is still $\displaystyle V_1 = \frac{kQ}{R}$

Also,
Potential due to the rod, at the center of the sphere, is given as,

$\displaystyle V_2 = \frac{kq}{L}ln\left(\frac{\sec\alpha+\tan\alpha}{\sec\alpha-\tan\alpha}\right)$

Since the sphere is grounded ,
$\displaystyle V_1+V_2 = 0$

$\displaystyle \frac{kQ}{R} + \frac{kq}{L}ln\left(\frac{\sec\alpha+\tan\alpha}{\sec\alpha-\tan\alpha}\right) = 0$

Note that,
$\displaystyle\tan\alpha = \frac{L}{2R}$ and $\displaystyle\sec\alpha = \frac{\sqrt{4R^2+L^2}}{2R}$

Substituting the values of $\displaystyle \tan\alpha$ and $\displaystyle \sec\alpha$ , we get,

$\displaystyle Q = -\frac{qR}{L}ln\left(\frac{\sqrt{4R^2+L^2}+L}{\sqrt{4R^2+L^2}-L}\right) \approx \boxed{-0.8813\mu C}$

Since the sphere is grounded, the electrostatic potential at every point on it and hence potential at center would be $0$ .

Say, charge $Q$ is induced.The induced charge would reside at the surface. Hence, the distance of center from every small charge $d q$

The potential due to induced charges at the center ( $V_{1}$ ) = $\displaystyle \int \frac{dq}{4\pi \epsilon_{0} R} = \frac{Q}{4 \pi \epsilon_{R}}$

Also, the potential at center due to rod ( $V_{2}$ ) = $\displaystyle 2 \int_{0}^{L/2} \frac{q/L dx}{4 \pi \epsilon_{0} \sqrt{R^2 + x^2}}$

Using $\frac{L}{2} = R$ , $V_{2} = \frac{q}{4 \pi \epsilon_{0} R} \ln(\sqrt{2} + 1)$

Hence, total potential at center = $V_{1} + V_{2} = 0$

Thus, $\frac{Q}{4 \pi \epsilon_{R}} + \frac{q}{4 \pi \epsilon_{0}R} \ln(\sqrt{2} + 1) = 0$

OR $Q = -q \ln(1 + \sqrt{2}) = -0.88 \mu C$