electric potential Begin Date: 2/8/2018 12:30:00 PM -Due Date: 2/152018 11:59:00
ID: 1872749 • Letter: E
Question
electric potential Begin Date: 2/8/2018 12:30:00 PM -Due Date: 2/152018 11:59:00 PM End Date: 31/2018 11:59:00 PM (17%) Problem 4: Charge 91 =-4.5 nC is located at the coordinate system origin, while charge 92-132 nC is located at (a, 0), where a = 1.3 m. The point P has coordinates (a, b), whereb 0.65 m. A third charge q312 nC will be placed later (a,b) qi Stheexpertta.com > 33% Part (a) Find the electric potential VP at point P. in volts. Assume the potential is zero at infinity Grade Summary 0% 100% sin0 tan() cos0 cotan asin acos0 atan acotanO sinh0 cosh0 tanh cotanhO Attempts remaining: 2 (3% per attempt) detailed view Degrees O Radians Hint I give up! Hints: 0% deduction per hint. Hints remaining: Feedback: 1% deduction per feedback. 46 33% Part (b) How much work W in joules, would you have to do to bring the third charge, q3. from very far away to the point P? 33% Part (c) What is the total potential energy U, in joules, of the final configuration of three charges?Explanation / Answer
(a)
Electric potential at point P,
Vp = kq1 / sqrt (a^2 + b^2) + kq2 / b
Vp = [9*10^9*-4.5*10^(-9) / sqrt (1.3^2 + 0.65^2)] + [9*10^9*-1.32*10^(-9) / 0.65]
Vp = -46.14 V
(b)
Work done to bring the third charge ,
W = q3*(Vp - Vinfinity)
W = 12*10^(-9)*(-46.14 - 0)
W = -553.7*10^(-9) J
(c)
Total potential energy = Total work done
work done to bring q1, W1 = 0
Work done to bring q2, W2 = q2*V1 = q2*kq1 / a
W2 = -1.32*10^(-9) * 9*10^9*-4.5*10^(-9) / 1.2
W2 = 41.12*10^(-9) J
U = W1 + W2 + W3
U = 0 + 41.12 - 553.7*10^(-9)
U = -512.56*10^(-9) J
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