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. Electric force experienced by `+1C` of charge at O due to chages at A and B will be of the same magnitude and in the same direction. The test charge +1C at O experiences force of repulsion due to `+3muC` and attraction due to `-3muC`.
The resultant force experienced by a unit charge is numerically equal to the electric field intensity at that point.
Hence `E_(0)=E_(OA)+E_(OB)`
`=2E_(OA),` because `E_(OA)=E_(OB)`
`=2xx(1/(4pi epsilon_(0)))q/((OA)^(2))`
`=(2xx9xx10^(9)xx3xx10^(-6))/((0.1)^(2))Vm^(-1)` (or `NC^(-1)`)
`=54xx10^(5)Vm^(-1)`
The resultant field at the midpoint is `5.4xx10^(6)Vm^(-1)`
b. If `+1C` is replaced by `-1.5xx10^(-9)C` then the force experienced by the new charge is `F=(E_(0))(q_(0))`
i.e. `F=-5.4xx10^(6)xx1.5xx10^(-9)N`
i.e. `F=-8.10xx10^(-3)N`
Hence the resultant force will act in the direction along BA which will be opposite to the direction of field at O.