Animated Solution for Physics - Electrostatics: Let a total charge $2Q$ be distributed in a sphere of radius $R$, with the charge density given by $\rho (r)=kr$, where $r$ is the distance from the centre. Two charges $A$ and $B$, of $-Q$ each, are placed on diametrically opposite points, at equal distance $a$, from the centre. If $A$ and $B$ do not experience any force, then

Imagine a solid sphere of radius $R$. It holds a total charge of $2Q$, but it's not uniformly distributed. The charge density, $\rho (r)$, increases linearly with distance as $kr$. Now, we place two negative charges, $-Q$ each, at points $A$ and $B$. These points are diametrically opposite, each at a distance $a$ from the center.

The problem states that charge $A$ experiences zero net force. How is that possible? Look closely at the forces acting on it. Charge $B$, being negative, will repel $A$ outwards. At the same time, the positive charge of the sphere enclosed within radius $a$ will attract $A$ towards the center. For equilibrium, these two opposing forces must perfectly balance each other.

First, we need to find the value of the constant $k$. We know the total charge is $2Q$. Let's consider a thin spherical shell of radius $r$ and thickness $dr$. Its volume is $4\pi r^{2}dr$. The charge in this shell, $dQ$, is density times volume.

Integrating this from $0$ to $R$ gives us the total charge.

Solving this integral, we find that $k$ equals $2Q$ divided by $\pi R^4$.

Next, let's find the electric field at distance $a$ from the center. We'll use Gauss's Law. We draw a Gaussian sphere of radius $a$. The charge enclosed is found by integrating our density function from $0$ to $a$.

Applying Gauss's Law, the electric field $E$ comes out to be $k$ times $a^2$ divided by $4{\epsilon }_0$.

Now, let's substitute the value of $k$ into our electric field expression. The field $E$ becomes $2Q{a}^2$ divided by $4\pi {\epsilon }_0{R}^4$. The attractive force on charge $A$, let's call it $F_1$, is simply its charge $Q$ multiplied by this electric field. This is the inward pull exerted by the inner positive sphere.

What about the outward push? Charge $B$ is at a distance of $2a$ from charge $A$. Using Coulomb's Law, the repulsive force $F_2$ between them is $\frac{1}{4\pi {\epsilon }_0}$ times $Q^2$ divided by the square of the distance, which is $(2a{)}^2$. This simplifies to $Q^2$ over $4\pi {\epsilon }_0(4a^2)$.

Finally, for charge $A$ to be in equilibrium, we equate the attractive force $F_1$ and the repulsive force $F_2$.

Notice how beautifully $Q^2$ and $4\pi {\epsilon }_0$ cancel out from both sides! Cross-multiplying gives us $8a^4=R^4$.

Taking the fourth root, we get our final answer: $a=8^{-1/4}R$. This problem is a brilliant fusion of Gauss's Law and Coulomb's Law. Think about this: what if the charge density was uniform instead of varying with $r$? The electric field would be proportional to $r$, not $r^2$, leading to a completely different equilibrium position. Always be prepared for such conceptual twists in JEE Advanced!