Question

A semicircle of radius a is in the first and second quadrants, with the center of curvature at the origin. Positive charge +Q is distributed uniformly around the left half of the semicircle, and negative charge -Q is distributed uniformly around the right half of the semicircle (in Fig.). What are the magnitude and direction of the net electric field at the origin produced by this distribution of charge?

Short answer

and direction of So the net electric field at the origin produced by this distribution of charge has the magnitude of ${\mathrm{E}}_{\mathrm{n}}=\frac{4\mathrm{kQ}}{{\mathrm{\pi a}}^2}$, and direction is + i

Step-by-step solution

Calculating differentia electric field

As shown in the figure the first arc is negative and the second is positive. The positive arc will produce an electric field in the 4th quadratic, and the negative arc will produce an electric field in the first quadratic at the same angle and magnitude so the vertical component of the -ve arc will cancel the vertical component of the positive and the horizontal components will be in the same direction and equal magnitudes.

The horizontal component of the net field is given by

${\mathrm{E}}_{\mathrm{n}}=2{\mathrm{E}}_{\mathrm{x}}$ (1)

The component ${\mathrm{E}}_{\mathrm{n}}$is given by

$\mathrm{E\_x}=\int \left(\mathrm{dE}\right)\mathrm{cos\theta }$ (2)

The differential electric field is given by

$dE=\frac{kdQ}{a^2}$ (3)

The distribution is uniform so

$\frac{\mathrm{Q}}{\mathrm{L}}=\frac{2\mathrm{Q}}{\mathrm{\pi a}}=\mathrm{\lambda }=\frac{\mathrm{dQ}}{\mathrm{dl}}$ (4)

Where: L is the arc length located in the 1st quadratic and $\mathrm{dl}=\mathrm{a}\left(\mathrm{d\theta }\right)$.

Substituting equations to get the magnitude and direction of the net electric field

Substitution from (4) in (3) gives

$\mathrm{dE}=\frac{\mathrm{k\lambda dl}}{{\mathrm{a}}^2}$ (5)

Substitution from (5) in (3) gives

$$\begin{gathered} E_x=\int \frac{k\lambda \cos \left(\theta \right)dI}{a^2} \\ ={\int }_a^{x/2}\frac{k\lambda \cos \left(\theta \right)ad\theta }{a^2} \\ =\frac{k\lambda }{a}{{\left[\sin \left(\theta \right)\right]}_0}^{\pi /2}=\frac{k\lambda }{a} \end{gathered}$$

Substitution in (1) gives

$E_n=\frac{2k\lambda }{a}=\frac{2k}{a}\frac{2Q}{a\pi }=\frac{4kQ}{\pi a^2}$

And the direction as explained n the first line is + i