Two point charges `q_(A)=3muC` and `q_(B)=-3muC` are located 0.2 m apart in vacuum.
a. What is the electric field at the mid point O of the line AB joining the two charges?
b. If a negative test charge of magnitude `1.5xx10^(-9)C` is placed at this point, what is the force experienced by the test charge?

Explanation

(a) The situation is represented in the given figure . O is the mid - point of line AB.
Distance between the two charges , AB =20 cm
`thereforeAO=OB=10cm`
Net electric field at point O=E
Electric field at point O caused by `+3muC` charge ,
`E_(1)=(3xx10^(-6))/(4piepsilon_(0)(AO)^(2))`
`=(3xx10^(-6))/(4piepsilon_(0)(10xx10^(-2))^(2))N//C` along OB
where , `epsilon_(0)=` Permittivity of free space
`(1)/(4piepsilon_(0))=9xx10^(9)Nm^(2)C^(-2)`
Magnitude of electric field at point O caused by `-3muC` charge.
`E_(2)=|(-3xx10^(-6))/(4piepsilon_(0)(OB)^(2))|`
`=(3xx10^(-6))/(4piepsilon_(0)(10xx10^(-2))^(2))N//C` along OB
`thereforeE=E_(1)+E_(2)`
`=2xx(9xx10^(9))xx(-3xx10^(-6))/((10xx10^(-2))^(2))`
[ Since the value of `E_(1)andE_(2)` are same , the value is multiplied with 2]
`=5.4xx10^(6)N//C` along OB
Therefore , the electric field at mid - point O is `5.4xx10^(6)NC^(-1)` along OB.
(b) A test chrge of amount `1.5xx10^(-9)C` is placed at mid - point O.
`q=1.5xx10^(-9)C`
Force experienced by the test charge =F
`thereforeF=qE`
`=1.5xx10^(-9)xx5.4xx10^(6)`
`=8.1xx10^(-3)N`
The force is directed along line OA . This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A. Therefore , the force experienced by the test charge is `8.1xx10^(-3)N` along OA.