homework_05_potential_energy_of_point_charges

Potential Energy of Point Charges

A point charge [q_2 = -3.2 \mu C] is fixed at the origin of a co-ordinate system as shown. Another point charge [q_1 = -0.4 \mu C] is is initially located at point [P], a distance [d_1 = 9.5 cm] from the origin along the [x]-axis

1)

What is [\Dleta U_{P R}], the change in potenial energy of charge [q_1] when it is moved from point [P] to point [R], located a distance [d_2 = 3.8 cm] from the origin along the [x]-axis as shown?

• Given

  • [\vec F_E = \frac{ 1}{ 4 \pi \varepsilon_0} \frac{ Q q}{ r^2}\hat{ r} \rightarrow \Delta U_{A B} = \frac{ Q q}{ 4 \pi \varepsilon_0} \left(\frac{ 1}{ r_B} - \frac{ 1}{ r_A}\right)]

• Let

  • [q_1 = 2.5 \mu C]
  • [q_2 = -3.2 \mu C]
  • [d_1 = 9.5 cm = .095 m]
  • [d_2 = 3.8 cm = .038 m]

• [\Delta U_{P R} = \frac{ q_1 q_2}{ 4 \pi \varepsilon_0} \left(\frac{ 1}{ d_2} - \frac{ 1}{ d_1}\right) = -1.17069]

2)

The charge [q_2] is now replaced by two charges [q_3] and [q_4] which each have a magnitude of [-1.6 \mu C], half of that of [q_2]. The charges are located a distance [a = 2.3 cm] from the oringin along the [y]-axis as shown. What is [\Delta U_{P R}], the change in potential energy now if charge [q_1] is moved from point [P] to point [R]?

• Given

  • [\vec F_E = \frac{ 1}{ 4 \pi \varepsilon_0} \frac{ Q q}{ r^2}\hat{ r} \rightarrow \Delta U_{A B} = \frac{ Q q}{ 4 \pi \varepsilon_0} \left(\frac{ 1}{ r_B} - \frac{ 1}{ r_A}\right)]

• Let

  • [q_1 = -0.4 \mu C]
  • [q_2 = -3.2 \mu C]
  • [q_3 = -1.6 \mu C]
  • [q_4 = -1.6 \mu C]
  • [d_1 = 9.5 cm = .095 m]
  • [d_2 = 3.8 cm = .038 m]
  • [a = 2.3 cm = .023 m]
  • [\theta_P = \arctan{ \left(\frac{ a}{ d_1}\right)}]
  • [\theta_R = \arctan{ \left(\frac{ a}{ d_2}\right)}]

• [\Delta U_{P R} = \left(\frac{ q_1 q_2}{ 4 \pi \varepsilon_0} \left(\frac{ 1}{ d_2}\right) \cos{ (\theta_R)\right)

  • \left(\frac{ q_1 q_2}{ 4 \pi \varepsilon_0} \left(\frac{ 1}{ d_1}\right) \cos{ (\theta_P)\right) = -.918373]
    • Somehow I got this problem correct even though I wasn't using the radius from using the similar side rather than the hypotenous of the triangle.

3)

What is the potential energy of the system composed of the three charges [q_1], [q_3], and [q_4], when [q_1] is at point [R]? Define the potential energy to be zero at infinity.

• Given

  • [\vec F_E = \frac{ 1}{ 4 \pi \varepsilon_0} \frac{ Q q}{ r^2}\hat{ r} \rightarrow \Delta U_{A B} = \frac{ Q q}{ 4 \pi \varepsilon_0} \left(\frac{ 1}{ r_B} - \frac{ 1}{ r_A}\right)]

• Let

  • [q_1 = -0.4 \mu C]
  • [q_2 = -3.2 \mu C]
  • [q_3 = -1.6 \mu C]
  • [q_4 = -1.6 \mu C]
  • [d_1 = 9.5 cm = .095 m]
  • [d_2 = 3.8 cm = .038 m]
  • [a = 2.3 cm = .023 m]
  • [\theta_P = \arctan{ \left(\frac{ a}{ d_1}\right)}]
  • [\theta_R = \arctan{ \left(\frac{ a}{ d_2}\right)}]
  • [h = \sqrt{ a^2 + d_2^2}]
  • [U_{R} = \sqrt{ \left(\left(\frac{ q_3 q_4}{ 4 \pi \varepsilon_0} \left(\frac{ 1}{ 2 a}\right)\right)
    • \left(\frac{ q_1 q_2}{ 4 \pi \varepsilon_0} \left(\frac{ 1}{ h}\right) \sin{ (\theta_R)}\right)\right)^2
  • \left(\left(\frac{ q_1 q_2}{ 4 \pi \varepsilon_0} \left(\frac{ 1}{ h}\right) \cos{ (\theta_R)}\right)\right)^2} = ]