Name: Electric Field of a Charged Plane Sheet
Uploaded: 2019-07-19
Duration: 6 min 58 s
Description: Another distribution of charge. This time on an infinite plane sheet. Let’s see once again how Gauss’s law comes to the rescue in electric field calculations.

Question

The figure shows two nonconducting spherical shells fixed in place. Shell $1$ has uniform surface charge density $+ 6.0 μ C m^{- 2}$ on its outer surface and radius $2$ ; shell $2$ has uniform surface charge density $+ 4.0 μ C m^{- 2}$ on its outer surface and radius $2.0 cm$ the shell centers are separated by $L = 12 cm .$ In unit-vector notation, what is the net electric field at $x = 2.0 cm ?$

Question Image

Accepted Answer

Given data is,

$+ 6.0 μ C m^{- 2}$

$σ_{2} = 4 μ C m^{- 2} = 4 \times 10^{- 6} C m^{- 2}$

$R_{1} = 3 cm = 3 \times 10^{- 2} m$

$R_{2} = 2 cm = 2 \times 10^{- 2} m$

Distance between shells is,

$L = 12 cm = 12 \times 10^{- 2} m$

$σ_{1}, σ_{2}$ are surface charge densities of shell $1$ and $2$ .

$R_{1}, R_{2}$ are radii of shell $1$ and $2$ .

Electric field at $x = 2 cm$ on $x$ -axis due to shell $1$ is equal to zero.

Since, the charge lies inside the spherical shell is zero.
$∴ E_{1} = 0$

Electric field at $x = 2 cm$ on $x$ -axis due to shell $2$ can be written as,

$E_{2} = \frac{1}{4 π ε_{0}} \frac{q_{2}}{{(L - x)}^{2}}$ (along negative $x$ -axis)

Here, $L - x$ is the distance between the point and shell $2$ ’ $L - x = 12 - 2 = 10 cm$

$q_{1} = σ_{2} (4 π R_{2}^{2})$ is the charge on shell $2$ .

$∴ E_{2} = \frac{1}{4 π ε_{0}} . \frac{σ_{2} 4 π R_{2}^{2}}{{(L - x)}^{2}} \Rightarrow E_{2} = \frac{σ_{2} R_{2}^{2}}{ε_{0} {(L - x)}^{2}}$

$E_{2} = \frac{4 \times 10^{- 6} {(2 \times 10^{- 2})}^{2}}{8 .85 \times 10^{- 12} {(10^{- 1})}^{2}} = \frac{4 \times 10^{- 6} \times 4 \times 10^{- 4}}{8 .85 \times 10^{- 12} \times 10^{- 2}}$

$E_{2} = \frac{6}{8 .85} \times 10^{4} \Rightarrow E_{2} = 1 .81 \times 10^{4} N C^{- 1}$

In vector form we can write,

${\vec{E}}_{2} = - 1.81 \times 10^{- 4} \hat{i} N C^{- 1}$

According to the superposition principle, the total electric field at $x = 2 cm$ can be written as,

$\vec{E} = {\vec{E}}_{1} + {\vec{E}}_{2} = 0 + (- 1 .81 \times 10^{4} \hat{i})$

$\vec{E} = - 1 .81 \times 10^{4} \hat{i} N C^{- 1}$

Resnick & Halliday Solutions for Chapter: Gauss' Law, Exercise 1: Problems

Resnick & Halliday Physics Solutions for Exercise - Resnick & Halliday Solutions for Chapter: Gauss' Law, Exercise 1: Problems

Questions from Resnick & Halliday Solutions for Chapter: Gauss' Law, Exercise 1: Problems with Hints & Solutions

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