Question

A charge $q_1=7\mu C$ is located at the origin and a second charge $q_2=-5\mu C$ is located on the x-axis, $0.30m$ from the origin. Find the electric field at the point $P$ which has coordinates $(0,0.40)m$

• A. $1.1\times {10}^5\hat{i}N/C$
• B. $2.5\times {10}^5\hat{j}N/C$
• C. $(1.1\times {10}^5\hat{i}+2.5\times {10}^5\hat{j})N/C$
• D. None of these

Answer 1

The correct option is C $(1.1\times {10}^5\hat{i}+2.5\times {10}^5\hat{j})N/C$


Diagram:
The total electric field $E$ at P equals to the vector sum $E_1$ and $E_2$, where $E_1$ is the field due to positive charge $q_1\left(7\mu C\right)$ at origin and $E_2$ is the field due to negative charge $(-5\mu C)q_2$.

Magnitude of fields $E_1$ and $E_2$ are to be calculated as follows:

$E_1=\frac{k|q_1|}{r_1{}^2}=9\times {10}^9\times \frac{7\times {10}^{-6}}{(0.40{)}^2}$

$=\frac{63\times {10}^3}{16}\times {10}^2$

$=3.93\times {10}^{5}N/C$

$E_2=\frac{k|q_2|}{r_2{}^2}=9\times {10}^9\times \frac{5\times {10}^{-6}}{(0.50{)}^2}$

$=1.8\times {10}^{5}N/C$

[From using Pythagoras theorem, distance $\left(r_2\right)$ between $q_2$ and point P, $r_2=\sqrt{(0.30{)}^2+(0.40{)}^2}=0.50m$]

Now, from the figure,

$\sin \theta =\frac{0.40}{0.50}=\frac{4}{5}$

$\cos \theta =\frac{0.30}{0.50}=\frac{3}{5}$

The vector $E_1$ has only $y$ component.

The vector $E_2$ has $x$ component given by $E_2\cos \theta =\frac{3}{5}{E}_2$ and a negative $y$ component given by $-E_2\sin \theta =\frac{-4}{5}{E}_2$.

Hence we can express vectors as:

$E_1=3.93\times {10}^5\hat{j}N/C$
$E_2=[1.08\hat{i}+(-1.44\hat{j}\left)\right]\times {10}^{5}N/C$

$\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}$

$=[1.08\times {10}^5\hat{i}+(3.9-1.44)\times {10}^5\hat{j}]N/C$

$=[1.1\times {10}^5\hat{i}+2.49\times {10}^5\hat{j}]N/C$

$=[1.1\times {10}^5\hat{i}+2.5\times {10}^5\hat{j}]N/C$