Question

Electric field is an important way of characterizing the electrical environment of a system of charges.
Two point charges $q_1$ and $q_2$ of magnitude $+{10}^{-8}C$ and $-{10}^{-8}C$ respectively are placed $0.1m$ apart. Calculate the electric fields at points $A$, $B$ and $C$ shown in the figure.

Answer 1

Electric field due to a point charge is given by:

$\overrightarrow{E}=\frac{q}{4{\epsilon }_o{r}^2}\hat{r}$ where $\hat{r}$ is unit vector from point charge to the point.

Using above formula with superposition principle gives:

${\overrightarrow{E}}_A=\frac{9\times {10}^9\times {10}^{-8}}{{0.05}^2}\hat{i}+\frac{9\times {10}^9\times {10}^{-8}}{{0.05}^2}\hat{i}$

$=7.2\times {10}^4\hat{i}N/C$

Similarly,

${\overrightarrow{E}}_B=-\frac{9\times {10}^9\times {10}^{-8}}{{0.05}^2}\hat{i}+\frac{9\times {10}^9\times {10}^{-8}}{{0.15}^2}\hat{i}$

$=-3.2\times {10}^4\hat{i}N/C$

For point C, using symmetry it can be concluded that vertical component of electric field is zero. Also, net electric field is twice the horizontal component of electric field due to each charge.

${\overrightarrow{E}}_C=2\times \frac{9\times {10}^9\times {10}^{-8}}{{0.1}^2}\cos \theta \hat{i}$

$=1.8\times {10}^{-3}\times \frac{0.05}{0.1}\hat{i}$ using definition of cosine

$=9\times {10}^3\hat{i}N/C$