We shall charge a capacitor from its initial potential difference V=Vi=0 to a final potential difference V=Vf.

Sunday, 26 August 2018

1:12 AM

Machine generated alternative text: Question: We shall charge a capacitor trom its initial potential difference V = Vi = 0 to a final potential difference V = Vf. part 1) How much work is done to add the first small amount ot charge dq to the capacitor (at Vi = 0) and now much work is done to add the last small* amount of charge dq to the capacitor (at V dWi dWf = (Incidentally, this question gives us some insight into the use ot calculus in the world. Here, dq is obviously not infinitesimal. The minimum charge transfer onto a capacitor is e, which is small, but not infinitesimal.) part 2) using the ideas in part 1 and the definition ot capacitance C, write an integral for the work done to charge the capacitor from uncharged to V = Vf. Your integral should be in terms of V and should not include dq. W = f OVf part 3) Hence write an expression for the potential energy stored on a capacitor with capacitance C charged to Vf. Part 4) Assume that the capacitor is a parallel plate capacitor with area A, plate separation d, negligible edge etfects and no dielectric. using your answer to pan 3), derive an expression for the energy per unit volume ot the vacuum between the plates. Give your answer in terms ottne electric tield E between the plates and do not include C in the expression; simplify it as much as possible. volume Check

 

Machine generated alternative text: Your answer is partially correct Feedback tor Part la) You have correctly answered this part Marks for this submissiom 0.20/0.20. Feedback tor Part 1b) The definition of electrical potential difference is dW = V dq so dWf = Vf dq. Marks for this submissiom 0.00/0.20. This submission attracted a penalty of 0.07. Feedback tor Part 2) W = dW = Vdq. aut trom the definitionotC, q = CVsodq Marks for this submissiom 0.00/0.20. This submission attracted a penalty of 0.07. Feedback tor Part 3) You have correctly answered this part Marks for this submissiom 0.20/0.20. Feedback tor Part 4) C = eoA/d. Electric field E = Vf/d, volume Ad. So (Ed)2 volume ¯ volume *This question gives us some insight into the use of calculus in the world. Here, dq is obviously not infinitesimal. The minimum charge transfer onto a capacitor is e, which is small, but not infinitesimal Because ottne uncertainty principle and quantisation, the same is true of (nearly?) all variables in the world. Marks for this submissiom 0.00/0.20. This submission attracted a penalty of 0.07. A correct answer is 0, which can be typed in as follows: A correct answer is Vf • dq, which can be typed in as follows: V f*dq A correct answer C • V • dV, which can be typed in as follows: A correct answer (C*V fA2)/2 , which can be typed in as follows: E , which can be typed in as follows: A correct answer

 

 

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