Question

A point charge q2 = -1.9 μC is fixed at the origin of a co-ordinate system as shown. Another point charge q1 = 4.5 μC is is initially located at point P, a distance d1 = 6.6 cm from the origin along the x-axis

1) What is ΔPE, the change in potenial energy of charge q₁ when it is moved from point P to point R, located a distance d₂ = 2.6 cm from the origin along the x-axis as shown?

2) The charge q2 is now replaced by two charges q3 and q4 which each have a magnitude of -0.95 μC, half of that of q2. The charges are located a distance a = 1.6 cm from the origin along the y-axis as shown. What is ΔPE, the change in potential energy now if charge q1 is moved from point P to point R?

3)What is the potential energy of the system composed of the three charges q1, q3, and q4, when q1 is at point R? Define the potential energy to be zero at infinity.

4)The charge q4 is now replaced by charge q5 which has the same magnitude, but opposite sign from q4 (i.e., q5 = 0.95 μC). What is the new value for the potential energy of the system?

5)Charges q3 and q5 are now replaced by two charges, q2 and q6, having equal magnitude and sign (-1.9μC). Charge q2 is located at the origin and charge q6 is located a distance d = d1 + d2 = 9.2cm from the origin as shown. What is ΔPE, the change in potential energy now if charge q1 is moved from point P to point R?

Answer 1

1.

r_i = initial distance between pair of charges = 6.6 cm = 0.066 m

r_f = final distance between pair of charges = 2.6 cm = 0.026 m

U_i = initial potential energy = k q₁ q₂/r_i = (9 x 10⁹) (4.5 x 10^-6) (- 1.9 x 10^-6)/(0.066) = - 1.2 J

U_f = final potential energy = k q₁ q₂/r_f = (9 x 10⁹) (4.5 x 10^-6) (- 1.9 x 10^-6)/(0.026) = - 2.96 J

change in potential energy = U = U_f - U_i = - 2.96 - (-1.2) = - 1.76 J

2)

r_i = initial distance between pair of charges = sqrt(6.6² + 1.6²) = 6.79 cm = 0.0679 m

r_f = final distance between pair of charges = sqrt(2.6² + 1.6²) = 3.05 cm = 0.0305 m

U_i = initial potential energy = k q₁ q₂/r_i = (9 x 10⁹) (4.5 x 10^-6) (- 1.9 x 10^-6)/(0.0679) = - 1.13 J

U_f = final potential energy = k q₁ q₂/r_f = (9 x 10⁹) (4.5 x 10^-6) (- 1.9 x 10^-6)/(0.0305) = - 2.52 J

change in potential energy = U = U_f - U_i = - 2.52 - (-1.13) = - 1.39 J

3)

U = - 2.52 + k q₃ q₄/r

U = - 2.52 + (9 x 10⁹) (- 0.95 x 10^-6) (- 0.95 x 10^-6)/(2 x 1.6) = - 2.3 J