InterviewSolution
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InterviewSolution
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Two point charges of `10^(-8)C and -10^(-8)C` are placed `0*1 m` apart. Calculate electric field intensity at A, B and C shown in Fig. |
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Answer» 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. |
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