1-A point charge q1 = 2.50 nC is placed at the origin, and a second point charge

Question

1-A point charge q1 = 2.50 nC is placed at the origin, and a second point charge q2 = 4.00 nC is placed on the x-axis at x = +29.00 cm. A third point charge q3 = 1.25 nC is to be placed on the x-axis between q1 and q2. (Take as zero the potential energy of the three charges when they are infinitely far apart.) (a) What is the potential energy of the system of the three charges if q3 is placed at x = +14.50 cm? J (b) Where should q3 be placed to make the potential energy of the system equal to zero? x = m

Explanation / Answer

Given,

q1 = 2.5 nC = 2.5 x 10-9 C placed at (0,0) ; q2 = -4 nC = -4 x 10-9 C ; at (+29 cm, 0 )

q3 = 1.25 nC = 1.25 x 10-9 C ;

(a) We need to find the potential energy of the system when, q3 ( +14.5 cm , 0).

The potential energy of the system will be:

U(sys) = U12 + U23 + U13

U(sys) = k q1q2/r12 + k q2q3/r23 + k q1q3/r13

U(sys) = 9 x 109 x 10-18 [ 2.5 x (-4) / 0.29 + (-4) x 1.25/(0.29-0.145) + 2.5 x 1.25 / 0.145 ]

U(sys) = 9 x 10-9 x (-34.5 + 34.5 + 21.6) = 194.4 x 10-9 J = 1.944 x 10-7 J

Hence, U(sys) =  = 1.944 x 10-7 J.

b)

U(sys) = U12 + U23 + U13 = 0

k q1q2/r12 + k q2q3/r23 + k q1q3/r13 = 0

9 x 109 x 10-18 [ 2.5 x (-4) / 0.29 + (-4) x 1.25/(0.29 - x) + 2.5 x 1.25 / x ] = 0

2.5 x (-4) / 0.29 + (-4) x 1.25/(0.29 - x) + 2.5 x 1.25 / x = 0

-34.5 - 5/(0.29 - x ) + 3.125 / x = 0

On simplifying we get a quadratic equation:

34.5 x2 - 11.875 x + 0.91 = 0

x = 0.22 and x = 0.91 m

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.