Two point charges `q_(1) and q_(2)`, of magnitude `+10^(–8) C and –10^(–8)C`, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C shown in Fig. 1.14.

Options

A. `E_(A)= 7.2xx10^(4)NC^(-1)` `E_(B)=3.2xx10^(4)NC^(-1)`
`E_(B)` is directed towards the RIGHT.`
B. `E_(A)= 7.2xx10^(-4)NC^(-1)` `E_(B)=3.2xx10^(4)NC^(-1)`
`E_(B)` is directed towards the left.`
C. `E_(A)= 7.2xx10^(4)NC^(-1)` `E_(B)=3.2xx10^(4)NC^(-1)`
`E_(B)` is directed towards the left.`
D. `E_(A)= 7.2xx10^(4)NC^(-1)` `E_(B)=3.2xx10^(-4)NC^(-1)`
`E_(B)` is directed towards the left.`

Answer

C

Explanation

The electric field vector E1A at A due to the positive charge q1 points towards the right and has a magnitude
`E_(1A)=((9xx10^(9)NmC^(2))xx(10^(-8)C))/((0.05m^(2)))=3.6cc10^(4)NC^(-1)`
The electric field vector E2A at A due to the negative charge q2 points towards the right and has the same magnitude. Hence the magnitude of the total electric field EA at A is
`E_(A)=E_(1A)+E_(2A)=7.2xx10^(4)NC^(-1)`
`E_(A)` is directed toward the light.
The electric field vector E1B at B due to the positive charge q1 points towards the left and has a magnitude
`E_(1B)=((9xx10^(9)Nm^(2)C^(-2))xx(10^(-8C)))/((0.05m)^(2))=3.6xx10^(4)NC^(-1)`
The electric field vector E2B at B due to the negative charge q2 points towards the right and has a magnitude
`E_(1B)=((9xx10^(9)Nm^(2)C^(-2))xx(10^(-8C)))/((0.15m)^(2))=4xx10^(3)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 directions in which these two vectors point are indicated in Fig. 1.14. The resultant of these two vectors is
`E_(C)=E_(1) cos (pi)/(3)+ E_(2) "Cos" (pi)/(3)=9xx10^(-3)NC^(-1)`
`E_(c)` points towards the right.