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.

Explanation

The charges `+ 10^(-8)C and -10^(-8)C and -10^(-8)C` are held at P and Q respectively, `PQ = 0*1m`
Also, `PA = PB = 0*05 m`
`QA = 0*05 and CP = CQ = 0*1m`
At A, field intensity due to charge `q_(1)`
`= (1)/(4pi in_(0)) (q_(1)xx1)/(AP^(2))` , along PA
`= (9xx10^(9)x10^(-8))/((0*05)^(2)) = 36000 N//C`
Field intensity due to charge `q_(2)`
`= (9xx10^(9)x10^(-8))/((0*05)^(2))` along `AQ = 36000 N//C`
`:.` Net field intensity at `A = 36000 + 36000`
`= 72000 N//C = 7*2xx10^(4) N//C` along AQ
At B. Field intensity due to charge `q_(1)`
`= (1)/(4pi in_(0)) (q_(1)xx1)/(PB^(2))` along PB producd
`(9xx10^(9)xx10^(-8))/((0*05)^(2)) = 4000 N//C`
`:.` Net field intensity at `B = 36000 - 4000`
`= 32000 N//c`
`3*2xx10^(4) N//C` , along PB produced
At C, Field intensity due to charge `q_(1)` ,
`E_(1) = (1)/(4pi in_(0)) (q)/(PC^(2))` , along PC.
`E_(1) = (9xx10^(9)xx10^(-8))/((0-1)^(2)) = 9xx10^(3) N//C`
Field intensity due to charge `q_(2)`
`E_(2) = (1)/(4pi in_(0)) (q_(2))/(QC^(2))` along CQ,
`E_(2) = (9xx10^(9)xx10^(-8))/((0*1)^(2)) = 9xx10^(3) N//C`
Both, `E_(1) and E_(2)` are equal in magnitude, acting at `60^(@)` to CR parallel to BA.
`:.` Resulant intensity at R
`= E_(1) cos 60^(@) + E_(2) cos 60^(@) = 2 E_(1) cos 60^(@)`
`= 2xx9xx10^(3)xx(1)/(2) = 9xx10^(3) N//C.`
It is represented by CR parallel to BA. The components of `vec(E_(1))` and `vec(E_(2))` in directions `_|_` to CR cancel out.