Question

The electric field intensity at a distance of 20 cm from the centre of a sphere is 10 V/m. The radius of the sphere is 5 cm. Determine the electric field intensity at a point 8 cm distance from the centre of the sphere.

Answer 1

Explanation:

Given,

E1 = 10V/m

r1 = 20 cm = 20 × 10^-2 m

r2 = 8 cm = 8 × 10^-2 m

r = 5 cm = 5 × 10^-2 m

q = ??

E2 = ??

E = 9×10^9 ( q /r^2 )

=> 10 = 9×10^9 ( q/400×10^-4 )

=> q = 4.44 × 10^-11 C

Now, electric field intensity at point 8 cm distance from the centre of the sphere is

E = 9×10^9 ( q /r^2 )

E = 9×10^9 ( 4.44 × 10^-11 ÷ 64 × 10^-4 )

E = 6.24 × 10^2 V/m

Answer 2

Answer:

A sphere of radius 5 cm carries a charge of $4.4*10^{-11}C$, the electric field at a distance of 8 cm from the center is $61.8V/m$

Explanation:

• The electric field is the electrostatic force on a unit charge due to a charge

• The electric field due to a sphere with surface charge $q$ and radius $R$ at a distance $r$ from the center of the sphere (outside), is given by the Gauss law as

                        $E_{out} =\frac{q}{4\pi \varepsilon _0r^{2} }$

  where the value of  $\frac{1}{4\pi \varepsilon _0} =9*10^{9}$

• electric field does not dependent on the dimension of the sphere

From the question, we have

at a distance  $r=20cm=0.20m$ from the center of the sphere the electric field is

$E_{out} =\frac{9*10^{9}*q }{(0.20)^{2} } =10V/m$

rearrange the equation to get charge on the sphere

$q=\frac{10(0.20^{2} )}{9*10^{9} } =4.4*10^{-11} C$

now the electric field at a distance $r=8cm=0.08m$ is

$E_{out} =\frac{9*10^{9}*4.4^{-11} }{(0.08)^{2} } =61.8 V/m$