A point charge q_1 = 5.50 nC is placed at the origin, and a second point charge

Question

A point charge q_1 = 5.50 nC is placed at the origin, and a second point charge q_2 = -4.50 nC is placed on the x-axis at x = +39.40 cm. A third point charge q_3 = 2.75 nC is to be placed on the x-axis between q_1 and q_2. (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 q_3 is placed at x = +19.70 cm? J (b) Where should q_3 be placed to make the potential energy of the system equal to zero? x = m

Explanation / Answer

here,

q1 = 5.5 * 10^-9 C is at x1 = 0

q2 = - 4.5 * 10^-9 C is at x2 = 0.394 m

q3 = 2.75 * 10^-9 C

a)

the potential energy of the system , PE = k * q1*q2/0.394 + k * q2*q3 /0.197 + k * q1*q3/0.197

PE = 9 * 10^9 * 10^-9 * 10^-9 * ( - 4.4 * 4.5 /0.394 - 4.5 * 2.75/0.197 + 5.5 * 2.75 /0.197)

PE = - 3.27 * 10^-7 J

b)

let the potential energy be zero when the charge q3 is placed at x

PE = k * q1*q2/0.394 + k * q2*q3 /(0.394 - x) + k * q1*q3/x

0 = ( - 4.4 * 4.5 /0.394 - 4.5 * 2.75/(0.394 - x) + 5.5 * 2.75 /x)

x = 0.15 m

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