Question:
A sphere of radius R has a volume density of charge $? = kr$, where r is the distance from the centre of the sphere and k is constant. The magnitude of the electric field which exists at the sur ace o the sphere is given by ($e_0$ = permittivity of the free space)
A sphere of radius R has a volume density of charge $? = kr$, where r is the distance from the centre of the sphere and k is constant. The magnitude of the electric field which exists at the sur ace o the sphere is given by ($e_0$ = permittivity of the free space)
Updated On: Sep 3, 2024
- $\frac{4\pi kR^{4}}{3\varepsilon_{0}}$
- $\frac{kR}{3\varepsilon_{0}}$
- $\frac{4\pi kR}{\varepsilon_{0}}$
- $\frac{kR^{2}}{4\varepsilon_{0}}$
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The Correct Option is D
Solution and Explanation
Given, $\rho=K \cdot r$
By Gauss's theorem
$E\left(4 \pi r^{2}\right) =\frac{\int \rho \times 4 \pi r^{2} d r}{\varepsilon_{0}} $
$=\frac{\int K r \times 4 \pi r^{2} d r}{\varepsilon_{0}} $
$\Rightarrow E =\frac{K r^{2}}{4 \varepsilon_{0}} $
Here $ r =R$
So, $E=\frac{K R^{2}}{4 \varepsilon_{0}}$
By Gauss's theorem
$E\left(4 \pi r^{2}\right) =\frac{\int \rho \times 4 \pi r^{2} d r}{\varepsilon_{0}} $
$=\frac{\int K r \times 4 \pi r^{2} d r}{\varepsilon_{0}} $
$\Rightarrow E =\frac{K r^{2}}{4 \varepsilon_{0}} $
Here $ r =R$
So, $E=\frac{K R^{2}}{4 \varepsilon_{0}}$
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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).