Question

3. Notice that if too much charge is placed on the capacitor in one step, this no longer corresponds to the physical situation of slowly charging the capacitor. For example, if you do it in two steps, then you would obtain an expression for the potential energy (49) U= с which is twice as large as the correct value. What is the value of U if all of the charge is placed on the capacitor in three steps? (Hint: add the second and third terms). 4. What is the value of U if all of the charge is placed on the capacitor in four steps? 5. Factor out all the common factors, and this will leave a sum. 6. Use the expression below to evaluate the sum, and write out an expression for the energy stored in the electric field in terms of Aq,c, and N. N2 + N N(N + 1) 2 2 j=1

Answer 1

are given work done in adding second charge War (AQ) ? after the second charge is adoled, net charge - sat sq hence С q V= 28q с Hence work done in adding third charge =oqu = 89%(209) 2 CAQ? C net hence work done Wi- W + Wz (82)? 2 (0972 + c (09)?[ 1 +27 - (0) 3 (02)21 0239 is added (4) After the third charge hat chargeq 304 hence work done in adding charge We a 84 x = 3 (02)2 a 309 Vs C eth c

value of u if charge is placed in four steps = Witwe (89) 2 [1+ 2) + 3 (09) с 11 (892 [1+2+3] –(i) U = 60q? Clearly from ean úr and (ii), if charge is added in N+1 steps then Us 04? [1+2+3+4+-+N] c N Aq2 C N(NH) (6) using the expression ja 2 j=1 2. (N?+N) sa N(NA) Sq we get 2c 2

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