Answer:
at y=6.29 cm the charge of the two distribution will be equal.
Explanation:
Given:
linear charge density on the x-axis, [tex]\lambda_1=8\times 10^{-6}\ C[/tex]
linear charge density of the other charge distribution, [tex]\lambda_2=-6\times 10^{-6}\ C[/tex]
Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.
Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.
we know, the electric field due to linear charge is given as:
[tex]E=\frac{\lambda}{2\pi.r.\epsilon_0}[/tex]
where:
[tex]\lambda=[/tex] linear charge density
r = radial distance from the center of wire
[tex]\epsilon_0=[/tex] permittivity of free space
Therefore,
[tex]E_1=E_2[/tex]
[tex]\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}[/tex]
[tex]\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}[/tex]
[tex]\frac{8\times 10^{-6}}{x} =\frac{6\times 10^{-6}}{0.11-x}[/tex]
[tex]x=0.0629\ m[/tex]
∴at y=6.29 cm the charge of the two distribution will be equal.