Saved Bookmarks
| 1. |
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 figure. |
|
Answer» Solution :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)=(9 xx 10^(9)) xx (10^(-8))/((0.05)^(2))` `=3.6 xx 10^(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_(1A)+ E_(2A) =7.2 xx 10^(4) NC^(-1)`. `E_(A)` is directed towards the right. 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) =(9xx 10^(9)) xx (10^(-8))/((0.05)^(2))=3.6 xx 10^(2) 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)=((9 xx 10^(9)) xx (10^(-8)))/((0.15)^(2)) =4 xx 10^(3) NC^(-1)` The magnitude of the total electric field at B is `E_(B)=E_(1B)-E_(2B)=3.2 xx 10^(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 The directions in which these two vectors point are indicated in figure. The resultant of these two vectors is `E_(C)=E_(1) cos "" pi/3+E_(2) cos ""pi/3=9 xx 10^(3) NC^(-1) E_(C)` points towards the right. |
|