em 5: Charge q--65 (10%) Problem 5: Charge q,--6.5 nC is located at the coordina
ID: 1398231 • Letter: E
Question
em 5: Charge q--65 (10%) Problem 5: Charge q,--6.5 nC is located at the coordinate system origin. while charge q2-3.4 nC is located at (a, 0), where a 0.75 m. The point P has coordinates (a, b), where b = 0.7 m. A third charge q3 = 10.5 nC will be placed later y (a,b) Randomized Variables 91 --6.5 nC q2 = 34 nC a= 0.75 m b=0.7 m 4310.5 nC qi 2 33% Part (a) Find the electric potential VP at point P, in volts. Assume the potential is zero at infinity. 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? Grade Summary Deductions Potential 0% 100% cos0 asin0 atan)acotan)sinh( coshO tanhO cotanhO Submissions Attempts remaining: 3 (3% per attempt) detailed view tan() ( acos) sin cotan) asin) 0 END Degrees O Radians DEL CLEAR Submit Hint I give up! Hints: 30% deduction per hint. Hints remaining: 2 Feedback: 30% deduction per feedback. D 33% Part (c) What is the total potential energy U, in joules, of the final configuration of three charges?Explanation / Answer
distance of P from q1 = r1 = sqrt(a^2+b^2) = sqrt
(0.75^2+0.7^2) = 1.02 m
distance of P from q2 = r2 = b = 0.75 m
potential at P due the charge q1 = v1 = k*q1/r1 =
-(9*10^9*6.5*10^-9)/1.02 = -57.35 v
potentail at P due to charge q2 = V2 = k*q2/r2 =
(9*10^9*3.4*10^-9)/0.7 = 43.71 V
part(A)
net potential at P = Vp = V1 + V2 = -13.64 V
part(B)
at far point potential = vf = 0
change in potential = dV = Vf - Vp = 13.64 V
work done = W = dV*q3 = 13.64*10.5*10^-9 = 1.43*10^-7 J
part(C)
potential energy = U = k*q1*q2/r12 + k*qq2*Q3/r23 + k*q1*Q3/r13
r12 = a = 0.75 m
r23 = b = 0.7 m
r13 = r1 = 1.02 m
U = -((9*10^9*6.5*10^-9*3.4*10^-9)/0.75) + ((9*10^9*3.4*10^-9*10.5*10^-9)/0.7) - ((9*10^9*6.5*10^-9*10.5*10^-9)/1.02)
u = -4.08*10^-7 J <<<<<<--------answer
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