Question

A uniform line charge extends from x = - 2.6 cm to x = + 2.6 cm and has a linear charge density of = 5.5 nC/m.(a) Find the total charge.Find the electric field on the y axis at the following distances.(b) y = 4 cm(c) y = 12 cm

Answer 1

Answer with Explanation:

We are given that

x=-2.6 cm to x=2.6 cm

Linear charge density=$\lambda=5.5nC/m=5.5\times 10^{-9} C/m$

$1nC=10^{-9} C$

Length of wire=$2.6-(-2.6)=2.6+2.6=5.2cm$

Length of wire=$\frac{5.2}{100}=0.052m$

1 m=100 cm

We know that

a.Linear charge density=$\frac{Q}{L}$

Where Total charge =Q

Length=L

Total charge,Q=$\lambda L$

Using the formula

Total charge,Q=$5.5\times 10^{-9}\times 0.052=2.86\times 10^{-10}C$

b.y=4 cm=$\frac{4}{100}=0.04m$

1 m=100 cm

Electric field on the y-axis is given by

$E=\frac{2\lambda}{4\pi\epsilon_0 y}(\frac{L}{\sqrt{4y^2+L^2}}$

$\frac{1}{4\pi\epsilon_0}=9\times 10^9 Nm^2/C^2$

Using the formula

$E=\frac{2\times 5.5\times 10^{-9}\times 9\times 10^9\times 0.052}{0.04\times \sqrt{4(0.04)^2+(0.052)^2}}$

$E=1348.8N/C$

c.y=12 cm=$\frac{12}{100}=0.12m$

Using the formula

$E=\frac{2\times 5.5\times 10^{-9}\times 9\times 10^9\times 0.052}{0.12\times \sqrt{4(0.12)^2+(0.052)^2}}$

$E=174.7N/C$