JEE Physics 2023 Question with Solution

Step-by-step solution

Standard Method

Given: Two isolated metallic solid spheres have radii $R$ and $2R$, and initially both have the same surface charge density $\sigma$.

Find: The ratio $\frac{\sigma'}{\sigma}$ for the bigger sphere after connecting the spheres by a thin conducting wire.

When the spheres are connected by a conducting wire, their potentials become equal. Therefore,

$$\frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_1'}{R} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_2'}{2R}$$

So,

$$Q_2' = 2Q_1'$$

Initially, the charges on the spheres are obtained from surface charge density:

$$Q_1 = \sigma \left(4 \pi R^2\right) = 4 \pi R^2 \sigma$$ $$Q_2 = \sigma \left(4 \pi (2R)^2\right) = 16 \pi R^2 \sigma$$

Hence total charge is

$$Q_1 + Q_2 = 20 \pi R^2 \sigma$$

Using conservation of charge,

$$Q_1' + Q_2' = Q_1 + Q_2$$

Substitute $Q_2' = 2Q_1'$:

$$Q_1' + 2Q_1' = 20 \pi R^2 \sigma$$ $$3Q_1' = 20 \pi R^2 \sigma$$ $$Q_1' = \frac{20 \pi R^2 \sigma}{3}$$

Therefore,

$$Q_2' = \frac{40 \pi R^2 \sigma}{3}$$

Now the new surface charge density of the bigger sphere is

$$\sigma' = \frac{Q_2'}{4 \pi (2R)^2}$$ $$\sigma' = \frac{\frac{40 \pi R^2 \sigma}{3}}{16 \pi R^2}$$ $$\sigma' = \frac{40}{3} \cdot \frac{1}{16} \sigma = \frac{5}{6} \sigma$$

Hence,

$$\frac{\sigma'}{\sigma} = \frac{5}{6}$$

Therefore, the correct option is D. The solution labels option C, but the worked result clearly gives $\frac{5}{6}$, which matches option D.

Common mistakes

• Using equal charge instead of equal potential after connecting the spheres. Connected conductors redistribute charge until their potentials are equal, not their charges. Set $\frac{Q_1'}{R} = \frac{Q_2'}{2R}$, not $Q_1' = Q_2'$.

• Calculating initial charge with radius instead of surface area. Surface charge density is charge per unit area, so use $Q = \sigma \cdot 4\pi r^2$. For the bigger sphere of radius $2R$, the area is $16\pi R^2$, not $8\pi R^2$.

• Using the wrong area for the final charge density of the bigger sphere. After redistribution, $\sigma' = \frac{Q_2'}{4\pi(2R)^2} = \frac{Q_2'}{16\pi R^2}$. Forgetting the square on $2R$ gives an incorrect ratio.