AL..E-field~~~－oob73zj92b的部落格

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AL..E-field~~~

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PHY Past paper MC 90 #28 95 #27 02 #20 以上3 題唔識 help @@ 更新: 90#28 有d 唔明.... 我咁計,計到 C = = 點解唔可以咁計gei ? @@ Work done = Increase in E.P.E = Qq/4piEb

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90. 28. This is quite a difficult one... The answer is E. qQ/ 4piE0 (1/b - 1/a) The outer surface is negatively charged due to induction. So, the potential at the surface of the inner sphere (take potential at infinity to be zero) = Q/ 4piE0 (1/b) - Q/ 4piE0 (1/a) As potential is the work done by external force per unit charge to bring a positive charge from infinity to the point considered. Hence, work done = qV = qQ/ 4piE0 (1/b - 1/a) 95. 27. The answer is C (1) and (2) only. Electric potential energy of the system = -Q2 / 4piE0r So, when r increases, the value increases as the value is less negative. So, (1) is right. At point D, the electric field due to +Q is not changed. Whereas for charge -Q, the magnitude of electric field is Q/4piE0r2 So, when r between -Q and point D decreases, the magnitude of electric field increases. So, (2) is correct. Electric potential at D due to +Q is not changed. For -Q, electric potential = -Q / 4piE0r, for r decreases, the value also decreases. So, this is wrong. 02.20. The answer is C. For A, there is no evidence showing whether q is travelling from X to Y or Y to X. For B, we see that between Q and q, there are attractions. So, they should have the opposite signs. For C, since they have the opposite signs, the electric potential energy between Q and q must be negative. And hence at X, the P.E. is more negative than that at Y. Energy is conserved. So, the kinetic energy at X must be higher than there at Y. So, at X, q has a higher speed than at Y. For D, at X, q is nearer to Q. And hence the electric potential is lower than that at Y. 2009-01-03 10:03:31 補充： Nope, you also need to consider the potential of the outer sphere. You haven't calculated it yet.

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