Answer to Question #93481 in Electricity and Magnetism for ROHIT SHARMA

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

a)     The electric flux Φ is given by formula

$Φ=4πr^2 (1)$

where E is the electric field on the surface, r is radius of a solid sphere

 

b)

Electric field at a distance of 0.50 m from the centre of the sphere

In this case, we find electric field inside solide sphere (r=0.75-0.5=0.25)

According to Gauss’s law, we can write

$4πr^2=ρ \frac {4}{3 \varepsilon} πr^3 (2)$

Using (3) we got

$E=\frac {ρr}{3 \varepsilon} (3)$

E=4.7 N/C

Electric field at a distance of 1.5 m from the centre of the sphere

In this case, we find electric field outside solide sphere (R=1.5-0.75=0.75)

$4πr^2=ρ \frac {4}{3 \varepsilon} πR^3 (4)$

Using (4) we got

$E=\frac {ρR^3}{3 \varepsilon r^2} (5)$

E=1.7 N/C

In the case, when 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.