Animated Solution for Physics - Electrostatics: Two spherical conductors $A$ and $B$ of radii $1\text{ mm}$ and $2\text{ mm}$ are separated by a distance of $5\text{ cm}$ and are uniformly charged. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres $A$ and $B$ is

Two Spheres, One Wire

Imagine two spherical conductors, let's call them $A$ and $B$, separated by a distance of $5\text{ cm}$. Sphere $A$ is the smaller one with a radius $r_A=1\text{ mm}$, while sphere $B$ is larger with a radius $r_B=2\text{ mm}$. Both are uniformly charged.

Now, we introduce a conducting wire and connect the two spheres. What happens next is a fundamental principle of electrostatics: charge will flow. It flows from the sphere at a higher potential to the one at a lower potential. This flow isn't endless; it stops the moment both spheres reach a state of equilibrium. In this state, their potentials become exactly equal. Let's denote this common potential as $V$. So, we have our first crucial equation:

Linking Field and Potential

Our goal is to find the ratio of the magnitudes of the electric fields at the surfaces of these two spheres. To do this, we need to connect the electric field ($E$) to the potential ($V$).

Recall the formula for the electric field at the surface of a conducting sphere of radius $r$ carrying charge $Q$:

And the potential at the surface is given by:

If we look closely, we can rewrite the electric field equation by pulling out the potential term. Notice that $\frac{kQ}{r^2}$ is just $\frac{1}{r}\times \left(\frac{kQ}{r}\right)$. Substituting $V$ into this gives us a beautiful, simplified relationship:

Inverse Proportions

Armed with this relationship, finding the ratio of the electric fields is a breeze. Let's set up the ratio for sphere $A$ and sphere $B$:

Because they share the same common potential $V$, it elegantly cancels out from the numerator and the denominator. This leaves us with an inverse relationship with their radii:

Now, it's just a matter of plugging in the given values. We know $r_B=2\text{ mm}$ and $r_A=1\text{ mm}$:

The units cancel out perfectly, giving us our final, pure ratio:

The electric field at the surface of the smaller sphere is twice as strong as the electric field at the surface of the larger sphere. This makes physical sense: a smaller radius means a sharper curvature, which leads to a higher concentration of charge and a stronger electric field.