Question

A continous line of charge of length $3d$ lies along the x-axis, extending from $x=+d$ to $x=+4d$. The lies carries a uniform linear charge density ${}^{\prime }\lambda$.
In terms of $d$, $\lambda$ and necessary constant, find the magnitude of the electric field at the origin.

• A. $\frac{\lambda }{5\pi {\epsilon }_{0}d}$
• B. $\frac{\lambda }{4\pi {\epsilon }_{0}d}$
• C. $\frac{3\lambda }{8\pi {\epsilon }_{0}d}$
• D. $\frac{3\lambda }{16\pi {\epsilon }_{0}d}$

Answer 1

The correct option is B $\frac{\lambda }{4\pi {\epsilon }_{0}d}$

Solution

A continuous line of charge of length $=3d$ lies along the X-axis

Extending from $x=+dtox=+4d$

The lines carries a uniform linear charge density $=\lambda$

The terms of $Q,\lambda$ and necessary constant

Finding the magnitude of the electric field at the origin$=?$

Solution

The electric field inline with and outside of actual charge distribution at a point $x=0$ can be found by integration of all the contributions of infinitesimal charges $dQ={\lambda }_{0}dx$ along the line charge,

i.e from $x=x_0$ to $x=\alpha$

$E\left(0\right)={\int }_d^{4d}dE\left(x\right)=={\int }_d^{4d}\frac{dQ}{4\pi E_0{x}^2}={\int }_d^{4d}\frac{{\lambda }_{0}dx}{4\pi E_0}=\frac{{\lambda }_0}{4\pi E_0}[-\frac{1}{x}{]}_d^{4d}=\frac{\lambda }{4\pi E_{0}d}$