Question

A thin metallic spherical shell of radius 50 cm has a total charge of 4.0 µC placed on it. At the center of the shell is placed a point charge of 2.00 µC. What is the electric field at a distance of 30.0 cm from the center of the spherical shell?

Answer 1

Answer:

The electric field is $6\times 10^{5}\ N/C$

Solution:

As per the question:

Radius of the spherical shell, R = 50 cm = 0.5 m

Charge on the spherical shell, Q = $4.0\mu C$

Charge placed at the center, Q' = $2.00\mu C$

Now, the electric field at a distance, r = 30.0 cm = 0.3 m is given by:

$\vec{E} = \vec{E_{point}} + \vec{E_{hsph}}$

where

$\vec{E_{point}}$ = electric field due to point charge

$\vec{E_{hsph}}$ = electric field inside the hollow sphere

$\vec{E} = \frac{1}{4\pi epsilon_{o}}. \frac{(Q + Q')}{r^{2}} + 0$

$\vec{E} = (9\times 10^{- 9})\times \frac{(2 + 4)\times 10^{- 6}}{0.3^{2}} + 0$

$\vec{E} =6\times 10^{5}\ N/C$

Answer 2

The magnitude of electric field at the center of the spherical shell is $6 \times 10^{5} \;\rm N/C$.

Given data:

The radius of spherical shell is, r = 50 cm = 0.50 m.

The total charge placed on shell is, $Q=4.0 \;\rm \mu C=4.0 \times 10^{-6} \;\rm C$.

The charge at center of shell is, $Q'=2.00 \;\rm \mu C=2.00 \times 10^{-6} \;\rm C$.

The field distance is, d = 30 cm = 0.30 m.

The region where the electrostatic force is experienced by particles is known as electric field.

The magnitude of total electric field is given as,

Total electric field = Electric field due to point charge + Electric field due to charge inside the sphere

$E=E' +E''\\\\E = \dfrac{kQ}{d^{2}}+\dfrac{kQ'}{d^{2}}$

Here, k is the Coulomb's constant.

Solving as,

$E = \dfrac{k}{d^{2}}(Q+Q')\\\\E = \dfrac{9 \times 10^{9}}{0.30^{2}}((4 \times 10^{-6})+(2 \times 10^{-6}))\\\\E = 6 \times 10^{5} \;\rm N/C$

Thus, the magnitude of electric field at the center of the spherical shell is $6 \times 10^{5} \;\rm N/C$.

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