When a $5 C$ charge is kept in a uniform electric field,a force of $5000 N$ acts on it. Find the potential difference between two points in that field,separated by a distance of $1 cm$. (in $V$)

An infinite number of electric charges, each equal to $5 \text{ nC}$ (magnitude), are placed along the $X$-axis at $x = 1 \text{ cm}, x = 2 \text{ cm}, x = 4 \text{ cm}, x = 8 \text{ cm}, \dots$ and so on. In this setup, if the consecutive charges have opposite signs, then the electric field in $\text{N/C}$ at $x = 0$ is: $\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\right)$

If the potential at the centre of a uniformly charged hollow sphere of radius $R$ is $V$,then the electric field at a distance $r$ from the centre of the sphere is $(r > R)$.

Four point positive charges of same magnitude $(Q)$ are placed at four corners of a rigid square frame of side $L$ as shown in the figure. The plane of the frame is perpendicular to the $Z$-axis. If a negative point charge $(-q)$ is placed at a small distance $z$ away from the center of the frame along the $Z$-axis $(z < < L)$, then:

As shown in the figure,at what distance in $cm$ from point $A$ will the electric field be zero?

Two non-conducting solid spheres of radii $R$ and $2R$,having uniform volume charge densities $\rho_1$ and $\rho_2$ respectively,touch each other. The net electric field at a distance $2R$ from the centre of the smaller sphere,along the line joining the centres of the spheres,is zero. The ratio $\frac{\rho_1}{\rho_2}$ can be;
$(A) -4$ $(B) -\frac{32}{25}$ $(C) \frac{32}{25}$ $(D) 4$