1. Setting up the Potential Equations

Let the induced charges on spheres A and B be $q_A$ and $q_B$. Since the spheres are neutral initially and isolated, conservation of charge gives:

$$ q_A + q_B = 0 \implies q_B = -q_A $$

We need to equate the potentials $V_A$ and $V_B$. The potential on the surface of sphere A is due to:

1. External charge $q$ at distance $l$.
2. Its own charge $q_A$ (radius $r$).
3. Charge on sphere B, $q_B$, at distance $a$.

Thus, potential at A ($V_A$):

$$ V_A = \frac{1}{4\pi\epsilon_0} \left[ \frac{q}{l} + \frac{q_A}{r} + \frac{q_B}{a} \right] $$

Similarly, potential at B ($V_B$) is due to charge $q$ (at distance $l+a$), its own charge $q_B$, and charge $q_A$ at distance $a$:

$$ V_B = \frac{1}{4\pi\epsilon_0} \left[ \frac{q}{l+a} + \frac{q_B}{r} + \frac{q_A}{a} \right] $$

2. Solving for Induced Charge

Since connected, $V_A = V_B$. We can cancel the common factor $\frac{1}{4\pi\epsilon_0}$ and substitute $q_B = -q_A$:

$$ \frac{q}{l} + \frac{q_A}{r} – \frac{q_A}{a} = \frac{q}{l+a} – \frac{q_A}{r} + \frac{q_A}{a} $$

Group terms involving $q$ on one side and $q_A$ on the other:

$$ \frac{q}{l} – \frac{q}{l+a} = – \frac{q_A}{r} + \frac{q_A}{a} – \frac{q_A}{r} + \frac{q_A}{a} $$ $$ q \left( \frac{1}{l} – \frac{1}{l+a} \right) = q_A \left( \frac{2}{a} – \frac{2}{r} \right) $$

Simplify the fractions:

$$ q \left( \frac{l+a – l}{l(l+a)} \right) = -2 q_A \left( \frac{1}{r} – \frac{1}{a} \right) $$ $$ q \frac{a}{l(l+a)} = -2 q_A \frac{a – r}{ar} $$

3. Applying Approximations

The problem states that the separation is much larger than the radius ($a \gg r$). Therefore, $a – r \approx a$.

$$ q \frac{a}{l(l+a)} \approx -2 q_A \frac{a}{ar} $$ $$ q \frac{a}{l(l+a)} \approx – \frac{2 q_A}{r} $$

Now, solve for the magnitude of $q_A$:

$$ q_A = – \frac{q r a}{2 l(l+a)} $$

The modulus of the charge is:

$$ |q_A| = |q_B| = \frac{q r a}{2 l(l+a)} $$