Question

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

(a) The situation is represented in the given figure. O is the mid-point of line AB.

Distance between the two charges, $AB=20cm$

$\therefore AO=OB=10cm$

Net electric field at point $O=E$

Electric field at point O caused by +3 C charge,

$E_1=\frac{3\times {10}^{-6}}{4\pi {\in }_0(AO{)}^2}=\frac{3\times {10}^{-6}}{4\pi {\in }_0(10\times {10}^{-2}{)}^2}N/C$ along OB

Where,

${\in }_0=$ Permittivity of free space

$\frac{1}{4\pi {\in }_0}=9\times {10}^{9}N{m}^2{C}^{-2}$

Magnitude of electric field at point O caused by - 3 C charge,

$E_2=\left|\frac{-3\times {10}^{-6}}{4\pi {\in }_0(OB{)}^2}\right|=\frac{3\times {10}^{-6}}{4\pi {\in }_0(10\times {10}^{-2}{)}^2}N/C$ along OB

$\therefore E=E_1+E_2$

$=2\times \left[\right(9\times {10}^9)\times \frac{3\times {10}^{-6}}{(10\times {10}^{-2}{)}^2}]$ [As $E_1$ = $E_2$, the value is multiplied with 2]

$=5.4\times {10}^6$ N/ C along OB

Therefore, the electric field at mid-point O is $5.4\times {10}^6$ N/ C along OB.

(b) A test charge of amount $q=1.5\times {10}^{-9}C$ is placed at mid-point O.

$q=1.5\times {10}^{-9}C$

Force experienced by the test charge $=F$

$\therefore F=qE$

$=1.5\times {10}^{-9}\times 5.4\times {10}^6$

$=8.1\times {10}^{-3}N$

The force is along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.

Hence, the force experienced by the test charge is $8.1\times {10}^{-3}N$along OA.