Chapter 24, Problem 083 Your answer is partially correct. Try again. In the gure

Question

Chapter 24, Problem 083 Your answer is partially correct. Try again. In the gure point p ls at a distance d -3.65 fro m partide l - 2e and distance d much work do we do and (c) what is the potential eneroy of the three-particle system? _ 1.12 m f om particle 2 Oz +2e), with both part les fixed in place. (a) with V-0 at infinity, what is V at p? tf we bring a particle of charge G 2e from infinity to h(b) now - (a) N (b) N (c) Number Click if you would like to Show Work for this question: Opan Shaw Wark Units

Explanation / Answer

The potential at point P will be the sum of individual potentials due to q1 and q2

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

Vp = k (q1/r1 + q2/r2)

Vp = 9 x 10^9 [-2e/3.65 + 2e/1.12]

Vp = 9 x 10^9 x 2 x 1.6 x 10^-19 [-1/3.65 + 1/1.12]

Vp = 1.78 x 10^-9 V

b)we know that,

U = q V

U = 2 e x 1.78 x 10^-9 = 2 x 1.6 x 10^-19 x 1.78 x 10^-9 = 5.696 × 10^-28 J

Hence, U = 5.696 × 10^-28 J

c)The U of the system is:

U = k q1q2/r12 + kq2q3/r23 + kq3q1/q31

r12 = sqrt (3.65^2 + 1.12^2) = 3.82 m

U = 9 x 10^9 x 4 x (1.6 x 10^-19)^2 (-1/3.82 + 1/1.12 - 1/3.65) = 3.29 x 10^-28 J

Hence, U = 3.29 x 10^-28 J

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