Surface Charge Density Ratio After Contact of Spheres

Learn how surface charge densities change when two conducting spheres are brought into contact and separated. This concept uses potential equalization

❓ Question

Two metal spheres of radii $R$ and $3R$ have surface charge density $\sigma$ .

If they are brought in contact and then separated, the surface charge density on the smaller and bigger sphere becomes:

\sigma_1 \text{ and } \sigma_2

respectively.

Find the ratio:

\frac{\sigma_1}{\sigma_2}

🖼 Question Image

Surface Charge Density Ratio After Contact of Spheres

✍️ Short Explanation

This problem is based on:

👉 Charge redistribution
👉 Conducting spheres in contact
👉 Surface charge density.

Main idea:

After contact:

V_1=V_2

for conducting spheres.

🔷 Step 1 — Initial Charges 💯

Surface charge density:

\sigma=\frac{Q}{4\pi R^2}

For smaller sphere:

Q_1=\sigma(4\pi R^2)

Q_1=4\pi\sigma R^2

For bigger sphere:

Radius:

3R

So:

Q_2=\sigma\left[4\pi(3R)^2\right]

Q_2=36\pi\sigma R^2

Total charge:

Q=40\pi\sigma R^2

🔷 Step 2 — Condition After Contact

For conducting spheres in contact:

V_1=V_2

Using:

V=\frac{kQ}{R}

So:

\frac{Q_1'}{R} = \frac{Q_2'}{3R}

Q_2'=3Q_1'

🔷 Step 3 — Use Charge Conservation

Q_1'+Q_2'=40\pi\sigma R^2

Substitute:

Q_2'=3Q_1'

4Q_1'=40\pi\sigma R^2

Q_1'=10\pi\sigma R^2

Q_2'=30\pi\sigma R^2

🔷 Step 4 — Find New Surface Charge Densities

For smaller sphere:

\sigma_1= \frac{Q_1'}{4\pi R^2}

\sigma_1=\frac{10\pi\sigma R^2}{4\pi R^2}

\sigma_1=\frac{5\sigma}{2}

For bigger sphere:

\sigma_2= \frac{Q_2'}{4\pi(3R)^2}

= \frac{30\pi\sigma R^2}{36\pi R^2}

\sigma_2=\frac{5\sigma}{6}

🔷 Step 5 — Required Ratio

\frac{\sigma_1}{\sigma_2} = \frac{5\sigma/2}{5\sigma/6}

=3

🔷 Step 6 — JEE Trap Alert 🚨

❌ Charges equal assume kar lena after contact

❌ Surface charge density aur charge confuse kar lena

❌ Bigger sphere ka area $9$ times hona bhool jaana

Remember: