Question

Consider two charged metallic spheres $${S}_{1}$$ and $${S}_{2}$$ of radii $${R}_{1}$$ and $${R}_{2}$$ respectively. The electric fields $${E}_{1}$$ (on $${S}_{1}$$) and $${E}_{2}$$ (on $${S}_{2}$$) on their surfaces are such that $${ E }_{ 1 }/{ E }_{ 2 }={ R }_{ 1 }/{ R }_{ 2 }$$. Then the ratio $${V}_{1}$$ (on $${S}_{1}$$)/$${V}_{2}$$ (on $${S}_{2}$$) of the electrostatic potentials on each sphere is:

Answer 1

Correct option is A. $${ \left( { R }_{ 1 }/{ R }_{ 2 } \right) }^{ 2 }$$

Let charge $$S_1 = Q_1$$

Charge on $$S_2 = Q_2$$

For $$S_1, E_1 = \dfrac{1}{4\pi \epsilon_0} \dfrac{Q_1}{R_1^2}$$

$$\Rightarrow Q_1 = 4\pi \epsilon_0 E_1 R_1^2$$ ....(i)

$$V_1 = \dfrac{1}{4\pi \epsilon_0} \dfrac{Q_1}{R_1}$$

$$\Rightarrow Q_1 = 4\pi \epsilon_0 V_1R_1$$ ...(ii)

Similarly for $$S_2 $$ $$Q_2 = 4\pi E_0 E_2R_2^2$$ ....(iii)

$$Q_2 = 4\pi \epsilon_0V_2R_2$$ ......(iv)

$$\dfrac{E_1}{E_2} = \dfrac{\dfrac{1}{4\pi \epsilon_0} \dfrac{Q_1}{R_1^2}}{\dfrac{1}{4\pi \epsilon_0} \dfrac{Q_2}{R_2^2}} = \dfrac{Q_1}{Q_2} \times \left(\dfrac{R_2}{R_1}\right)^2$$

Given $$\dfrac{E_1}{E_2} = \dfrac{R_1}{R_2} \Rightarrow \dfrac{Q_1}{Q_2}\times \left(\dfrac{R_2}{R_1}\right)^2 = \dfrac{R_1}{R_2}$$

$$\Rightarrow \dfrac{Q_1}{Q_2} = \left(\dfrac{R_1}{R_2}\right)^3$$ ...(v)

$$\Rightarrow \dfrac{V_1}{V_2} = \dfrac{\dfrac{1}{4\pi \epsilon_0} \dfrac{Q_1}{R_1}}{\dfrac{1}{4\pi \epsilon_0} \dfrac{Q_2}{R_2}} = \dfrac{Q_1}{Q_2} \times \dfrac{R_2}{R_1} = \left(\dfrac{R_1}{R_2}\right)^3 \times \dfrac{R_2}{R_1}$$

$$\Rightarrow \dfrac{V_1}{V_2} = \left(\dfrac{R_1}{R_2}\right)^2$$