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Two point charges A = +3 nC and B = +1 nC are placed 5 cm apart in air. The work done to move charge B towards A by 1 cm is |
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Answer» `1.35xx10^(-7)J` `B=+1 nC = 1xx10^(-9)C` Distance, `r_(1)=5 cm = 0.05 m = 5XX10^(-2)m` WORKDONE `(W)=U_(B)-U_(A)` `= (kq_(1)q_(2))/(r_(2))-(kq_(1)q_(2))/(r_(1))` Here, `r_(2)=r_(1)-1` `r_(2)=5-1=4 cm = 0.04 m = 4xx10^(-2)m` `= kq_(1)q_(2)[(1)/(r_(2))-(1)/(r_(1))]` `= 9xx10^(9)xx3xx10^(-9)xx1xx10^(-9)` `= [(1)/(4xx10^(-2))-(1)/(5xx10^(-2))]` `=(27xx10^(-9)xx1)/(5xx4xx10^(-2))` `= (27)/(20)xx10^(-7)` `= 1.35xx10^(-7)J` |
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