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(B) The electric field $E$ at a distance $r$ $(r \ge R)$ from the centre of a charged sphere is given by $E = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac

Vedclass · Accepted Answer

$R \sqrt{\frac{\sigma}{\epsilon_{0} E}}$

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(B) The electric field $E$ at a distance $r$ $(r \ge R)$ from the centre of a charged sphere is given by $E = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q}{r^{2}}$.Surface charge density $\sigma$ is defined as $\sigma = \frac{q}{4 \pi R^{2}}$,which implies $q = 4 \pi R^{2} \sigma$.Substituting the value of $q$ into the electric field equation:$E = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{4 \pi R^{2} \sigma}{r^{2}}$$E = \frac{R^{2} \sigma}{\epsilon_{0} r^{2}}$Rearranging for $r^{2}$:$r^{2} = \frac{R^{2} \sigma}{\epsilon_{0} E}$Taking the square root on both sides:$r = R \sqrt{\frac{\sigma}{\epsilon_{0} E}}$.

$A$ metal sphere of radius $R \ cm$ is charged with $4 \pi \mu C$ and is situated in air. If $\sigma$ is the surface charge density and $E$ is the electric intensity at a distance $r$ from the centre of the sphere,then $r$ is equal to ($\epsilon_{0}$ is the permittivity of free space).

Similar Questions

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