Two spheres of radius a and b respectively are charged and joined by a wire. ratio of electric field

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1. Electric Charges and Fields

normal

Two spheres of radius $a$ and $b$ respectively are charged and joined by a wire. The ratio of electric field of the spheres is

$a/b$

$b/a$

${a^2}/{b^2}$

${b^2}/a$

Solution

(b) Joined by a wire means they are at the same potential. For same potential $\frac{{k{Q_1}}}{{{a_1}}} = \frac{{k{Q_2}}}{{{a_2}}}$ $==>$ $\frac{{{Q_1}}}{{{Q_2}}} = \frac{a}{b}$
Further, the electric field at the surface of the sphere having radius $R$ and charge $Q$ is $\frac{{kQ}}{{{R^2}}}.$
$\frac{{{E_1}}}{{{E_2}}} = \frac{{k{Q_1}/{a^2}}}{{k{Q_2}/{b_2}}} = \frac{{{Q_1}}}{{{Q_2}}} \times \frac{{{b^2}}}{{{a^2}}} = \frac{b}{a}$

Standard 12

Physics

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Two spheres of radius $a$ and $b$ respectively are charged and joined by a wire. The ratio of electric field of the spheres is

Solution

Similar Questions

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