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A point charge of +10 mu C is placed at a distance of 20 cm from another identical point charge of +10 muC. A point charge of -2mu C is moved from point a to b as shown in the figure . Calculate the change in potential energy of the system ? Interpret your result . |
Answer» Solution : `q_(1)= 10muC= 10xx10^(-6)C` `q_(2) =- 2 MUC = -2 xx10^(-6)C ` DISTANCE ,`r = 5cm = 5xx10^(-2)` m Change in potential energy `DeltaU = (9xx10^(9) xx10^(2)xx10^(6)XX(-2xx10^(-6)))/(5xx10^(-2)) ` `=-36xx10^(9)xx10^(-12)xx10^(2)= -36xx10^(-1)` `DeltaU = 3.6 J` Negative sign implies that to move the charge `-2muC` no external work is required. SYSTEM spends its STORED energy to move the charge from point A to point < `DeltaU=-3.6 J` negtive sign implies that to move the charge `-2muC` no external work is required system spend its stored energy to movethe charge from poitnt a to point b. |
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