**Question:**

Consider a sphere of radius $R$ which carries a uniform charge density $\rho$. If a sphere of radius $\frac{\mathrm{R}}{2}$ is carved out of it, as shown, the ratio $\frac{\left|\overrightarrow{\mathrm{E}}_{\mathrm{A}}\right|}{\left|\overrightarrow{\mathrm{E}}_{\mathrm{B}}\right|}$ of magnitude of electric field $\overrightarrow{\mathrm{E}}_{\mathrm{A}}$ and $\overrightarrow{\mathrm{E}}_{\mathrm{B}}$, respectively, at points $\mathrm{A}$ and $\mathrm{B}$ due to the remaining portion is:

Correct Option: 2

**Solution:**

### Related Questions

Three charges $+Q, q,+Q$ are placed respectively, at distance, $\mathrm{d} / 2$ and d from the origin, on the $x$-axis. If the net force experienced by $+Q$, placed at $x=0$, is zero, then value of $\mathrm{q}$ is:

Charge is distributed within a sphere of radius $R$ with a volume charge density $\mathrm{p}(\mathrm{r})=\frac{\mathrm{A}}{\mathrm{r}^{2}} \mathrm{e}^{-2 \mathrm{r} / \mathrm{a}}$ where $\mathrm{A}$ and $\mathrm{a}$ are constants. If $Q$ is the total charge of this charge distribution, the radius $\mathrm{R}$ is:

Two identical conducting spheres $A$ and $B$, carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is F. A third identical conducting sphere, $\mathrm{C}$, is uncharged. Sphere $\mathrm{C}$ is first touched to $A$, then to $B$, and then removed. As a result, the force between $A$ and $B$ would be equal to

Shown in the figure are two point charges $+Q$ and $-Q$ inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If $\sigma_{1}$ is the surface charge on the inner surface and $Q_{1}$ net charge on it and $\sigma_{2}$ the surface charge on the outer surface and $\mathrm{Q}_{2}$ net charge on it then :

Two charges, each equal to $\mathrm{q}$, are kept at $\mathrm{x}=-\mathrm{a}$ and $\mathrm{x}=\mathrm{a}$ on the $x$-axis. A particle of mass $m$ and charge $q_{0}=\frac{q}{2}$ is placed at the origin. If charge $\mathrm{q}_{0}$ is given a small displacement $(y<<a)$ along the $y$-axis, the net force acting on the particle is proportional to

Two balls of same mass and carrying equal charge are hung from a fixed support of length $l$. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, $x$ between the balls is proportional to :

Two identical charged spheres suspended from a common point by two massless strings of length $l$ are initially a distance $d(d<<l)$ apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity $v$. Then as a function of distance $x$ between them,

A charge $Q$ is placed at each of the opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the net electrical force on $Q$ is zero, then $Q / q$ equals:

If $g_{E}$ and $g_{M}$ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan’s oil drop experiment could be performed on the two surfaces, one will find the ratio

Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges on them repel each other with a force $F$ when kept apart at some distance. A third spherical conductor having same radius as that $B$ but uncharged is brought in contact with $B$, then brought in contact with $C$ and finally removed away from both. The new force of repulsion between $B$ and $C$ is