Two point charges `q_(A) = 3 mu C` and `q_(B) = -3 muC` are located 20 cm apart in vaccum (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 the 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
`therefore`AO = OB = 10 cm
Net electric field at point O = E
Electric field at point O caused by `+3 muC` charge,
`E_(1) = (3 xx 10^(-6))/(4 pi epsilon_(0)(AO)^(2))=(3 xx 10^(-6))/(4piepsilon_(0)(10 xx 10^(-2))^(2))N//C` along OB
Where,
`epsilon_(0)` = Permittivity of free space
`(1)/(4 piepsilon_(0)) = 9 xx 10^(9) Nm^(2)C^(2)`
Magnitude of electric field at point O caused by `−3muC` charge,
`E_(2) = |(-3 xx 10^(-6))/(4 pi epsi_(0)(OB)^(2))| = ( 3 xx 10^(6))/(4 pi epsi_(0)( 10 xx 10^(-2))^(2))N//C` along OB
`therefore E= E_(1) + E_(2)`
`= 2 xx [(9 xx 10^(9)) xx (3 xx 10^(6))/(10 xx 10^(-2))^(2))]` ["Since the values of `E_(1)` and `E_(2)` are same, the value is multiplied with 2]
`= 5.4 × 10^(6) N//C` along OB
Therefore, the electric field at mid-point O is `5.4 × 10^(6) N C^(−1)` along OB.
(b) A test charge of amount`1.5 × 10^(−9)` C is placed at mid-point O.
`q = 1.5 × 10^(−9) C`
Force experienced by the test charge = F
`thereforeF = qE`
`= 1.5 × 10^(−9) × 5.4 × 10^(6)`
`= 8.1 × 10^(-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 `8.1 × 10^(−3) N` along OA.