Question:
If an insulated non-conducting sphere of radius $R$ has charge density $ \rho . $ The electric field at a distance r from the centre of sphere $(r > R)$ will be:
If an insulated non-conducting sphere of radius $R$ has charge density $ \rho . $ The electric field at a distance r from the centre of sphere $(r > R)$ will be:
Updated On: May 22, 2024
- $\frac{\rho R}{3 \varepsilon_{0}}$
- $\frac{\rho r}{\varepsilon_{0}}$
- $\frac{\rho r}{3 \varepsilon_{0}}$
- $\frac{3 \rho R}{\varepsilon_{0}}$
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The Correct Option is C
Solution and Explanation
Charge which is enclosed in the surface is given by
$\sigma=\frac{4}{3} \pi r^{3} \rho$
Now according to Gaussian law
$\varepsilon \,\phi\, E d s=\sigma,$
$ \varepsilon_{0} E 4 \pi r^{2}=\frac{4}{3} \pi r^{3} \rho $
$E=\frac{\rho r}{3 \varepsilon_{0}}$
$\sigma=\frac{4}{3} \pi r^{3} \rho$
Now according to Gaussian law
$\varepsilon \,\phi\, E d s=\sigma,$
$ \varepsilon_{0} E 4 \pi r^{2}=\frac{4}{3} \pi r^{3} \rho $
$E=\frac{\rho r}{3 \varepsilon_{0}}$
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).