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 $\frac{V_{1} (on S_{1})}{V_{2} (on S_{2})}$ of the electrostatic potentials on each sphere is:

(\frac{R_{1}}{R_{2}})

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{(\frac{R_{1}}{R_{2}})}^{2}

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(\frac{R_{2}}{R_{1}})

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{(\frac{R_{1}}{R_{2}})}^{3}

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Solution

The correct option is B ${(\frac{R_{1}}{R_{2}})}^{2}$
Electric field at a point on the surface of the sphere is given by

$∴ E = \frac{ρ R^{3}}{3 ϵ_{0}}$

$Let ρ_{1} and ρ_{2} are the charge densities of two sphere .$

$E_{1} = \frac{ρ R_{1}}{3 ε_{0}} and E_{2} = \frac{ρ_{2} R_{2}}{3 ε_{0}}$

$∵ \frac{E_{1}}{E_{2}} = \frac{ρ_{1} R_{1}}{ρ_{2} R_{2}} = \frac{R_{1}}{R_{2}}$

$This gives ρ_{1} = ρ_{2} = ρ$

Potential at a point on the surface of the sphere

$V = \frac{1}{4 π ε_{0}} \frac{Q}{R}$

$V = \frac{ρ R^{3}}{3 ε_{0} R} ⎛ ⎜ ⎜ ⎜ ⎝ ∵ ρ = \frac{Q}{\frac{4}{3} π R^{3}} ⎞ ⎟ ⎟ ⎟ ⎠$

$V = \frac{ρ R^{2}}{3 ε_{0}} So, V_{1} = \frac{ρ R_{1}^{2}}{3 ε_{0}} V_{2} = \frac{ρ R_{2}^{2}}{3 ε_{0}}$

$∴ \frac{V_{1}}{V_{2}} = {(\frac{R_{1}}{R_{2}})}^{2}$

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Consider two charged metallic spheres S1 and S2 of radii R1 and R2, respectively. The electric fields E1 (on S1) and E2 (on S2) on their surfaces are such that E1/E2=R1/R2. Then the ratio V1(on S1)V2(on S2) of the electrostatic potentials on each sphere is: