The electric field vector `E_(1A)` at A due to the positive charge `q_(1)` points towards the right and has a magnitude
`E_(1A)=((9xx10^(9)Nm^(2) C^(-2))xx(10^(-8)C))/((0.05 m)^(2))=3.6xx10^(4) NC^(-1)`
The electric field vector `E_(2A)` at A due to the negative charge `q_(2)` points towards the right and has the same magnitude. Hence the magnitude of the total electric field `E_(A)` at A is
`E_(A)=E_(1A)+E_(2A)=7.2xx10^(4) NC^(-1)`
`E_(A)` is directed towards the right. ltbtgt The electric field vector `E_(1B)` at B due to the positive charge `q_(1)` points towards the left and has a magnitude.
`E_(1B)=((9xx10^(9) Nm^(2) C^(-2))xx(10^(-8) C))/((0.05 m)^(2))=3.6xx10^(4) NC^(-1)`.
the electric field vector `E_(2B)` at B due to the negative charge `q_(2)` points towards the right and has a magnitude.
`E_(2B)=((9xx10^(9) Nm^(2)C^(-2))xx(10^(-8) C))/((0.15 m)^(2))=4xx10^(3) NC^(-1)`.
The magnitude of the total electric field at B is `E_(B)=E_(1B)-E_(2B)=3.2 xx10^(4) NC^(-1)`.
The magnitude of the total electric field at B is `E_(B)=E_(1B)-E_(2B)=3.2xx10^(4) NC^(-1)`.
`E_(B)` is directed towards the left.
the magnitude of each electric field vector at point C, due to charge `q_(1)` and `q_(2)` is
`E_(1C)=E_(2C)=((9xx10^(9) Nm^(2)C^(-2))xx(10^(-8) C))/((0.10 m)^(2))=9xx10^(3) NC^(-1)`
The directions in which these two vectors point are indicated in Fig. The redultant of these two vector is `E_(C)=E_(1) "cos" pi/3+E_(2) "cos"pi/3 =9xx10^(3) NC^(-1)`
`E_(C)` point towards the right.
