Electrostatics

Given, $q_1=q_2=8 \propto \mathrm{C}=8 \times 10^{-6} \mathrm{C}$ and side of triangle, $x=0.2 \mathrm{~m}$

Since, electric potential, $V=\frac{k q}{r}$

$\therefore$ Potential at third point, $V_3=\frac{k q_1}{r}+\frac{k q_2}{r}$

$\Rightarrow V_3=\frac{9 \times 10^9 \times 8 \times 10^{-6}}{0.2} \times 2=7.2 \times 10^5 \mathrm{~V}$

As we know that according to Gauss's law,

$\oint \mathbf{E} \cdot d s=\frac{q_{\text {enc. }}}{\varepsilon_0}$

Hence, Gauss's law is used to determine electric field due to symmetric charge distribution inside the Gaussian surface.

Given, charge on outer sphere = q

Inner sphere is grounded.

Since, charge flows till potential (V) becomes same.

and $V=\frac{kq}{r}$

where, $r$ is radius of sphere $\left(r_2 > r_1\right)$.

$\begin{aligned} \frac{k q^{\prime}}{r_1} & =\frac{-k q}{r_2} \\ \Rightarrow \quad q^{\prime} & =-q\left(\frac{r_1}{r_2}\right)\end{aligned}$

As we know that,

Electric field lines emerge from the positive charge and converge to negative charge. Tangent drawn at any point on the field line gives the direction of electric field at that point.

Electric field lines do not intersect each other. If they do, so there will be two directions of force at a single point which is not possible.

Hence, statement (d) is incorrect.