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Unformatted text preview: David McEwen S0091331 1 ENTA11005 - Assignment 4 Question 1 (i) For the force between the two charges to remain the same after Q 1 is halved and they are separated by a factor of 4, Q 2 must obviously increase in charge. r Q Q F 2 1 = By using the above equation and making the mentioned changes the change in Q 2 can be found:- 1 2 2 1 2 1 2 1 8 8 4 2 4 2 Q rF Q rF Q Q rF Q Q F r Q Q = = × = × = × Therefore, for the force between the charges to remain the same Q 2 equals the product of 8 times the distance between the charges times force all divided by the charge of Q 1 (ii) If a conducting rod is earthed while held closely to a negatively charged object then the conducting rod would become positively charged. The golf-leafed electroscope shown in the picture is negatively charged so the gold leaves are separated because the negative charges are repelling each other. If the conducting rod which is now positively charged touched the golf-leafed electroscope which is negatively charged then the gold leaves would become closer together. The positive charge of the rod negatively charged then the gold leaves would become closer together....
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- Spring '08
- Capacitance, Electric charge, conducting rod, David McEwen, Hm -1