Answer with Explanation:
We are given that
x=-2.6 cm to x=2.6 cm
Linear charge density=[tex]\lambda=5.5nC/m=5.5\times 10^{-9} C/m[/tex]
[tex]1nC=10^{-9} C[/tex]
Length of wire=[tex]2.6-(-2.6)=2.6+2.6=5.2cm[/tex]
Length of wire=[tex]\frac{5.2}{100}=0.052m[/tex]
1 m=100 cm
We know that
a.Linear charge density=[tex]\frac{Q}{L}[/tex]
Where Total charge =Q
Length=L
Total charge,Q=[tex]\lambda L[/tex]
Using the formula
Total charge,Q=[tex]5.5\times 10^{-9}\times 0.052=2.86\times 10^{-10}C[/tex]
b.y=4 cm=[tex]\frac{4}{100}=0.04m[/tex]
1 m=100 cm
Electric field on the y-axis is given by
[tex]E=\frac{2\lambda}{4\pi\epsilon_0 y}(\frac{L}{\sqrt{4y^2+L^2}}[/tex]
[tex]\frac{1}{4\pi\epsilon_0}=9\times 10^9 Nm^2/C^2[/tex]
Using the formula
[tex]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}}[/tex]
[tex]E=1348.8N/C[/tex]
c.y=12 cm=[tex]\frac{12}{100}=0.12m[/tex]
Using the formula
[tex]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}}[/tex]
[tex]E=174.7N/C[/tex]
