Question

The volume charge density of a solid sphere of radius 0.75 m is 0.25 nCm−3 . Apply Gauss’s law to calculate the
a) electric flux through the sphere and
b) electric field at a distance of 1.5 m and at a distance of 0.50 m from the centre of the sphere, respectively. If this sphere were made of conducting material, what would the values of the electric fields be at these distances?

Answer 1

Question #86382, Physics / Electromagnetism

Electric field inside and outside of uniformly charged solid sphere:

Radius of the charged solid sphere (R = 0.75 m)

Volume charged density of solid sphere ($\rho = 0.25\, nC/m^3$)

Electric charge on sphere (Q)

$$Q = \rho V$$

Where $V$ is volume of solid sphere i.e. $\left(\frac{4}{3}\pi R^3\right)$

$Q =$ Total charge enclosed by the sphere.

(a) Electric flux through the sphere

Flux = $\oint E \, ds$

$$\begin{array}{l} = E \cdot A \quad \text{Where } A = \text{electric field passing through the surface} \\ = E \cdot 4 \cdot \pi \cdot R^2 \quad E = \text{Electric field on the surface}. \end{array}$$

(Note: We can find electric field by substituting radius of solid sphere in the formula of electric field inside the solid sphere as given below)

(b) Electric field inside the solid sphere i.e. (radius of Gaussian sphere = 0.5 m)

$$\oint E \, ds = \frac{Q'}{\epsilon}$$

$\oint E \cdot ds \cdot \cos 90{}^\circ = \frac{Q'}{\epsilon}$ Since E and ds are perpendicular to each other.

$$E \cdot 4\pi r^2 = \rho \cdot \frac{4}{3}\pi r^3 / \varepsilon$$ $$E = \frac{\rho r}{3\varepsilon}$$

On substituting values

$$E = \frac{0.25 \cdot 0.5 \cdot 10^{-9}}{3 \cdot 8.85 \cdot 10^{-12}} \quad N/C$$ $$E = 0.0047 \cdot 10^3 \quad \text{N/C}$$

Electric field outside the solid sphere i.e. (radius of Gaussian sphere = 1.5

$$\oint E \, ds = \frac{Q}{\epsilon}$$

$\oint E \cdot ds \cdot \cos 90{}^\circ = \frac{Q}{\epsilon}$ Since E and ds are perpendicular to each other.

$$E \cdot 4\pi r^2 = \rho \cdot \frac{4}{3}\pi R^3 / \varepsilon$$ $$E = \frac{\rho R^3}{3\varepsilon r^2}$$ $$E = 0.0017 \cdot 10^3 \quad \text{N/C}$$

(c) If this sphere were made of conducting material then

Electric field inside the conductor will be zero and outside it will be same as the electric field outside the solid sphere.

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