A +3.0-µC point charge is initially extremely far from a positive point charge Q.You find that it takes 41 J of work to bring the +3.0-µC charge to the point x = 3.0 mm,y = 0.0 mm and the point charge Q to the point x = -3.0 mm,y = 0.00 mm.What is Q? (k = …)

Question

Quizplus · Accepted Answer

We can use the formula for electric potential energy:

U = k*q1*q2/r

where U is the potential energy, q1 and q2 are the charges, r is the separation distance, and k is Coulomb's constant (k = 9.0 x 10^9 N*m^2/C^2).

Since the +3.0-µC charge is initially extremely far from the other charge Q, we can assume that their initial potential energy is zero. As the +3.0-µC charge is brought to a position 3.0 mm away from Q, the potential energy changes by an amount of 41 J, so we can write:

41 J = k*(3.0 µC)*Q/0.006 m

Solving for Q gives:

Q = (41 J*0.006 m)/(k*3.0 µC)
  = (41 J*0.006 m)/(9.0 x 10^9 N*m^2/C^2 * 3.0 x 10^-6 C)
  = 9.1 µC

Therefore, the answer is B) 9.1 µC.

A +30-µC Point Charge Is Initially Extremely Far from a Positive