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- nbsp- nbsp-

Electric field at a point on the surface of the sphere is given by ∴ E =ρ R^33 ϵ_0 Let ρ_1 and ρ_2 are the charge densities of two sphere. E_1 =ρ R_13ε_0 an ...

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Standard X

Physics

Electric Potential and Potential Difference

Consider two ...

Question

Consider two charged metallic spheres $S_1~ \text{and}~ S_2$ of radii $R_1~ \text{and}~ R_2$, respectively. The electric fields $E_1~ (\text{on}~ S_1) ~\text{and}~ E_2 ~(\text{on} ~S_2)$ on their surfaces are such that $E_1/E_2 = R_1/R_2$. Then the ratio $\dfrac{V_1(\text{on}~ S_1)}{ V_2(\text{on}~ S_2)}$ of the electrostatic potentials on each sphere is:

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Solution

Electric field at a point on the surface of the sphere is given by

$\therefore E =\dfrac{\rho R^3}{3~\epsilon_0}$

$\text{Let} ~\rho_1 ~\text{and}~\rho_2~ \text{are the charge densities of two sphere}.$

$E_1 =\dfrac{\rho R_1}{3\varepsilon_0} ~\text{and}~E_2 =\dfrac{\rho_2 R_2}{3\varepsilon_0}$

$\because ~\dfrac{E_1}{E_2} =\dfrac{\rho_1 R_1}{\rho_2 R_2} =\dfrac{R_1}{R_2}$

$\text{This gives}~\rho_1 = \rho_2 = \rho$

Potential at a point on the surface of the sphere

$V = \dfrac{1}{4 \pi \varepsilon _0 }\dfrac{Q}{R}$

$ V =\dfrac{\rho R^3}{3 \varepsilon_0R}\left (\because \rho =\dfrac{Q}{\dfrac{4}{3}\pi R^3} \right)$

$V =\dfrac{\rho R^2}{3\varepsilon
_0}\\ \text{So,}\\~V_1 =\dfrac{\rho R^2_1}{3 \varepsilon _0} \\~V_2 =\dfrac{\rho R^2_2}{3 \varepsilon_0}$

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

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