homework_01_point_charges_in_two_dimensions

Point Charges in Two Dimensions

Three charges ([q_1 = 5.9 \mu C, q_2 = -4.3 \mu C, and q_3 = 3.6 \muC]) are located at the vertices of an equilateral triangle with side [d = 9.5 cm] as shown.

1)

What is [F_{3,x}], the value of the [x]-component of the net force on [q_3]?

• [F_{ij} = \kappa \frac{ q_i q_j}{ r^2} \hat r]
  • [ \kappa = \frac{ 1}{ 4 \pi \epsilon_0} = 8.99 \times 10^9 \frac{ Nm^2}{ C^2} ]
  • [ \mu = 10^{-6}]
• Let:
  • [ i = q_1 = 5.9 \mu C]
  • [ j = q_2 = -4.3 \mu C ]
  • [ k = q_3 = 3.6 \mu C ]
  • [ d = 0.095 m ]
  • [ \theta = 60 ]
• [ F_{k,x} = \kappa \frac{ q_i q_k}{ d^2} \cos{ \left( \theta \right)} +\kappa \frac{ q_j q_k}{ d^2} \cos{ \left( \theta \right)} = 18.2888 N]

2)

What is [F_{3,y}], the value of the [y]-component of the net force on [q_3]?

• [F_{ij} = \kappa \frac{ q_i q_j}{ r^2} \hat r]
  • [ \kappa = \frac{ 1}{ 4 \pi \epsilon_0} = 8.99 \times 10^9 \frac{ Nm^2}{ C^2} ]
  • [ \mu = 10^{-6}]
• Let:
  • [ i = q_1 = 5.9 \mu C]
  • [ j = q_2 = -4.3 \mu C ]
  • [ k = q_3 = 3.6 \mu C ]
  • [ d = 0.095 m ]
  • [ \theta = 60 ]
• [ F_{k,y} = \kappa \frac{ q_i q_k}{ d^2} \sin{ \left( \theta \right)} - \kappa \frac{ q_j q_k}{ d^2} \sin{ \left( \theta \right)} = 4.96896 N]

3)

A charge [q_4 = 3.6 \mu C] is now added as shown.

What is [F_{2,x}], the [x]-component of the new net force on [q_2]?

• [F_{ij} = \kappa \frac{ q_i q_j}{ r^2} \hat r]
  • [ \kappa = \frac{ 1}{ 4 \pi \epsilon_0} = 8.99 \times 10^9 \frac{ Nm^2}{ C^2} ]
  • [ \mu = 10^{-6}]
• Let:
  • [ i = q_1 = 5.9 \mu C]
  • [ j = q_2 = -4.3 \mu C ]
  • [ k = q_3 = q_4 3.6 \mu C ]
  • [ d = 0.095 m ]
  • [ a = \left( \frac{ d}{ 2} \right) = .0475 m]
  • [ b = \sqrt{d^2 - a^2} = .0822]
  • [ \theta = \arctan{ \left( \frac{ a}{ b} \right)} = 30 ]
• [ F_{j,x} = \kappa \frac{ q_i q_j}{ d^2} +2\left( \kappa \frac{ q_k q_j}{ d^2} \sin{ \left( \theta \right)} \right) = -40.6916 N]

4)

What is [F_{2,y}], the [y]-component of the new net force on [q_2]?

• [F_{ij} = \kappa \frac{ q_i q_j}{ r^2} \hat r]
  • [ \kappa = \frac{ 1}{ 4 \pi \epsilon_0} = 8.99 \times 10^9 \frac{ Nm^2}{ C^2} ]
  • [ \mu = 10^{-6}]
• Let:
  • [ i = q_1 = 5.9 \mu C]
  • [ j = q_2 = -4.3 \mu C ]
  • [ k = q_3 = q_4 3.6 \mu C ]
  • [ d = 0.095 m ]
  • [ a = \left( \frac{ d}{ 2} \right) = .0475 m]
  • [ b = \sqrt{d^2 - a^2} = .0822]
  • [ \theta = \arctan{ \left( \frac{ a}{ b} \right)} = 30 ]
• [F_{j,y} = 0 ]

5)

What is [F_{1,x}], the [x]-component of the new net force on [q_1]?

• [F_{ij} = \kappa \frac{ q_i q_j}{ r^2} \hat r]
  • [ \kappa = \frac{ 1}{ 4 \pi \epsilon_0} = 8.99 \times 10^9 \frac{ Nm^2}{ C^2} ]
  • [ \mu = 10^{-6}]
• Let:
  • [ q_1 = 5.9 \mu C]
  • [ q_2 = -4.3 \mu C ]
  • [ q_{3,4} = q_3 = q_4 = 3.6 \mu C ]
  • [ r = 0.095 m ]
  • [ a = \left( \frac{ r}{ 2} \right) = .0475 m]
  • [ b = \sqrt{r^2 - a^2} = .0822]
  • [ \theta = \arctan{ \left( \frac{ a}{ b} \right)} = 30 ]
• [ F_{q_1, x} = \kappa \frac{ q_2 q_1}{ r^2} - 2 \left(\kappa \frac{ q_{3,4} q_1}{ r^2} \sin{ \left(\theta \right)}\right) = 4.11398 N]
  • Remember to include [\mu] when substituting for [q_i].
    • Rember to include [\mu^2] when substuting for [q_j q_i].

6)

How would you change [q_1] (keeping [q_2], [q_3] and [q_4] fixed) in order to make the net force on [q_2] equal to zero?

• Decrease its magnitude and change its sign.
  • Let:
    • [ i = q_1 = 5.9 \mu C]
    • [ j = q_2 = -4.3 \mu C ]
    • [ k = q_3 = q_4 3.6 \mu C ]
  • [F_{ij} > 2F_{kj} ]