Animated Solution for Physics - Electrostatics: 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 centre of the spheres, is zero. The ratio ${\rho }_1/{\rho }_2$ can be

Let's visualize the setup. We have two non-conducting solid spheres touching each other. Let the smaller sphere be $S_1$ with a radius $R$ and centre $C_1$. Let the larger sphere be $S_2$ with a radius $2R$ and centre $C_2$. Because they are touching, the distance between their centres is simply $R+2R=3R$.

The question tells us that the net electric field is zero at a distance of $2R$ from the centre of the smaller sphere, along the line joining their centres. If we move $2R$ from the centre of the smaller sphere, we can either go towards the larger sphere, reaching a point $P$, or go away from it, reaching a point $Q$. Let's analyze both cases to find the possible ratios of their charge densities.

The Point Between the Spheres

Let's start with Case 1, where the point $P$ lies between the two centres. $P$ is at a distance of $2R$ from the smaller sphere's centre $C_1$. Since the total distance between the centres is $3R$, $P$ must be at a distance of $3R-2R=R$ from the larger sphere's centre $C_2$.

There is a catch here: $P$ is outside the smaller sphere, but it lies inside the larger sphere!

Now, let's write the electric field at $P$. For the smaller sphere, $P$ is an outside point, so we use the formula for a point charge at the centre. The field $E_1$ is:

For the larger sphere, $P$ is an inside point at a distance $R$ from its centre. Using the formula for the field inside a solid sphere ($E=\frac{kQr}{R^3}$), $E_2$ is:

For the net electric field at $P$ to be zero, the fields from both spheres must cancel each other out. This means they must be equal in magnitude and opposite in direction. Equating $E_1$ and $E_2$, we get:

Simplifying this, we find that $Q_2=2Q_1$.

Now, let's express the charges in terms of their volume charge densities. Charge is density times volume. So, $Q_1={\rho }_1\left(\frac{4}{3}\pi R^3\right)$, and $Q_2={\rho }_2\left(\frac{4}{3}\pi (2R{)}^3\right)=8{\rho }_2\left(\frac{4}{3}\pi R^3\right)$.

Substituting these into our relation $Q_2=2Q_1$, we get:

This gives us our first possible ratio.

The Point Outside Both Spheres

Now let's move to Case 2, where the point $Q$ is on the left side of the smaller sphere. $Q$ is at a distance of $2R$ from $C_1$. The distance from the larger sphere's centre $C_2$ will be $2R+3R=5R$. In this case, point $Q$ lies outside both the spheres.

Since $Q$ is outside both spheres, we can treat both of them as point charges concentrated at their respective centres. The electric field $E_1$ due to the smaller sphere is:

The electric field $E_2$ due to the larger sphere is:

For the net electric field at $Q$ to be zero, the vector sum of $E_1$ and $E_2$ must be zero. Since they are on the same side, their fields must have opposite signs to cancel out. Adding them and equating to zero, we get:

The negative sign indicates that the charge densities must have opposite signs.

Finally, let's substitute the charges with their densities and volumes just like before:

Plugging these in, the volumes cancel out, and we get:

By analyzing both possible locations for the point of zero electric field, we found two valid ratios for the charge densities. Therefore, the ratio $\frac{{\rho }_1}{{\rho }_2}$ can be either $4$ or $-\frac{32}{25}$.