Question

Two point charges $q_A=3\mu C$ and $q_B=–3\mu C$ are located 20 cm apart in vacuum. What is the electric field at the midpoint O of the line AB joining the two charges?

If a negative test charge of magnitude $1.5\times {10}^{–9}C$ is placed at this point, what is the force experienced by the test charge?

Answer 1

AB=20cm



AO=OB=10cm=0.1m
$q_A=3\mu C=3\times {10}^{-6}C{q}_B=-3\mu C=-3\times {10}^{-6}C$
The electric field at a point due to a charge q is E= $\frac{1}{4\pi {\epsilon }_0}.\frac{q}{r^2}$
Where, r is the distance between charge and the point.
Electric field due to $q_A$ at O is $E_A.$
$E_A=\frac{1}{4\pi {\epsilon }_0}.\frac{q}{(AO{)}^2}{E}_A=\frac{9\times {10}^9\times 3\times {10}^{-6}}{(0.1{)}^2}=\frac{27\times {10}^3}{0.1\times 0.1}=2.7\times {10}^{6}N/C$
The direction of $E_A$ is A to O i.e., towards O or towards OB as the electric field is always directed away from positive charge.
Electric field due to $q_B$ at O is $E_B$
$E_B=\frac{1}{4\pi {\epsilon }_0}.\frac{q_B}{(OB{)}^2}{E}_A=\frac{9\times {10}^9\times 3\times {10}^{-6}}{(0.1{)}^2}=\frac{27\times {10}^3}{0.1\times 0.1}=2.7\times {10}^{6}N/C$
The direction of $E_B$ is O to B i.e., towards O or towards OB as the electric field is always directed towards the negative charge.
Now, we see that both $E_{A}and{E}_B$ are in same direction. So, if the resultant electric field at O is E,
$E=E_A+E_B=2.7\times {10}^6+2.7\times {10}^6=5.4\times {10}^{6}N/C$
The direction of E (resultant electric field) will be from O to B or towards B.

Let us consider, the charge q is placed at the mid-point O. According to the question,



$q=-1.5\times {10}^{-9}C$
By the basic definition of electric field,
$E=\frac{F}{q}$
or F=qE, where, E is the net electric field at point O.
$F=-1.5\times {10}^{-9}\times 5.4\times {10}^6=-8.1\times {10}^{-3}N$
The direction of force is opposite to the direction of field because the charge q is negatively charged. Thus, the direction of force is from O to A.