Question

CALC 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 (Fig. P21.86). What are the magnitude and direction of the net electric field at the origin produced by this distribution of charge? 

Answer 1

Electric filed at the origin:

Like charges repel each other.

Unlike charges attract each other.

Sin component will cancel each other.

\(d E_{\text {Total }}=2 d E \cos \theta\)

electric filed due to small portion of charge dQ:

\(d E_{\text {Total }}=2 d E \cos \theta\)

electric filed due to small portion of charge \(\mathrm{dQ}\) :

\(d E=\frac{k(d Q)}{a^{2}}\)

\(d Q=\lambda a d \theta\)

\(\lambda=\) charge density

\(d E_{\text {Total }}=2 d E \cos \theta\)

\(E_{\text {Total }}=\int_{0}^{\frac{\pi}{2}} 2 \frac{k \lambda a}{a^{2}} \cos \theta d \theta\)

From,

\(\int_{0}^{\frac{\pi}{2}} \cos \theta d \theta=[\sin \theta]_{0}^{\frac{\pi}{2}}=1\)

\(E_{\text {Total }}=\frac{2 k \lambda a}{a^{2}}\)

\(Q=\lambda a \int_{0}^{\frac{\pi}{2}} d \theta=\lambda a \frac{\pi}{2}\)

\(\lambda a=\frac{2 Q}{\pi}\)

The magnitude of electric field at origin.

\(E_{\text {Total }}=\frac{4 Q k}{\pi a^{2}}\)

The direction of the electric field at origin.

The net field is along the positive \(x\) -axis.

\(E_{\text {Total }}=\frac{4 Q k}{\pi a^{2}} \hat{\imath}\)

Answer 2

As the left half is positive and right half is negative

electric field along vertical direction would cancel and horizontal direction would add

Answer 3

Answer 4

Answer 5