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Two point charges 2 muC and 0.1 muC are held at a separation of 20 mm. What work is required by the external agent to move the charges to half of the given separation? |
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Answer» Solution :Here, Work done = Final potential energy - Initial potential energy `rArr W=(1)/(4pi epsilon_(0)) (q_(1)q_(2))/(r.)-(1)/(4pi epsilon_(0)) (q_(1)q_(2))/(r )` Here, `q_(1)=2muC= 2XX10^(-6)C` `q_(2)=0.1muC=0.1 XX 10^(-6)C` `r.=` Final separation between `q_(1) and q_(2)` `=10 m m=10xx10^(-3)m` r = Initial separation between `q_(1) and q_(2)` `=20 mm=20xx10^(-3)m` `therefore W=(q_(1)q_(2))/(4pi epsilon_(0)) [(1)/(r.)-(1)/(r )]` `=9xx10^(9)xx2xx10^(-6)xx0.1xx10^(-6)` `[(1)/(10xx10^(-3))-(1)/(20xx10^(-3))]` `=9xx10^(9)xx2xx10^(-6)xx0.1xx10^(-6)xx(1,000)/(20)` `=9xx10^(-2)J` |
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