homework_02_electric_field_from_point_charges

Electric Field from Point Charges

Two point charges [q_1 = -4.1 \mu C] and [q_2 = 7.7 \mu C] are fixed along the [x]-axis, separated by a distance [ r = 0.095 m]. Point [P] is located at [(x,y) = (d,d)].

• Let
  • [q_1 = -4.1 \mu C]
  • [q_2 = 7.7 \mu C]
  • [d = 0.095 m]
  • [P = (d, d)]
• Given
  • [E = k \frac{ Q}{ r^2}]
    • For point charges

1)

What is [E_x( P)], the value of the [x]-component of the electric field produced by [q_1] and [q_2] at point [P]?

• Let
  • [r = [\sqrt{ 2 d^2}]
  • [\theta = 45]
• [E_x( P) = k \frac{ q_1}{ r^2} \cos{ (\theta)} = k \frac{ q_1}{ d^2} \cos{ (\theta)} = -1.44395 \times 10^6]

2)

What is [E_y( P)], the value of the [y]-component of the electric field produced by [q_1] and [q_2] at point [P]?

• [E_x( P) = k \left(\frac{ q_1}{ 2d^2} \sin{ (\theta)}\right) + \frac{ q_2}{ d^2} = 6.22619 \times 10^6]
  • remember [q_1] is negative.

3)

A third point charge [q_3 = 2.3 \mu C] is now positioned along the [y]-axis at a distance [d = 0.095 m] from [q_1] as shown. What is [E_x( P)], the [x]-component of the field produced by all three charges at point [P]?

• Let

  • [q_1 = -4.1 \mu C]
  • [q_2 = 7.7 \mu C]
  • [q_3 = 2.3 \mu C]
  • [d = 0.095 m]
  • [P = (d, d)]

• Given

  • [E = k \frac{ Q}{ r^2}]
    • For point charges

• [ E_x( P) = k \left(\frac{ q_1}{ 2d^2} \cos{ (\theta)} + \frac{ q_3}{ d^2}\right) = 847133]

4)

Suppose all charges are now doubled (i.e., [\hat q_1 = 2 q_1 = -8.2 \mu C], [\hat q_2 = 2 q_2 = 15.4 \mu C], [\hat q_3 = 2 q_3 = 4.6 \mu C]), how will the electric field at [P] change?

• Its magnitude will double and it s direction will remain the same.
  • [E( P) = \sqrt{ \left(k \left(\frac{ q_1}{ 2d^2} \cos{ (\theta)}
    • \frac{ q_3}{ d^2}\right)\right)^2
    • \left(k \left(\frac{ q_1}{ 2d^2} \sin{ (\theta)}
    • \frac{ q_2}{ d^2}\right)\right)^2}] [= \sqrt{ \left(k \left(\frac{ q_1}{ 2d^2} \cos{ (\theta)}
    • \frac{ q_3}{ d^2}\right)\right)^2
    • \left(k \left(\frac{ q_1}{ 2d^2} \sin{ (\theta)}
    • \frac{ q_2}{ d^2}\right)\right)^2}]
  • [\hat E( P) = \sqrt{ \left(k \left(\frac{ 2 q_1}{ 2d^2} \cos{ (\theta)}
    • \frac{ 2 q_3}{ d^2}\right)\right)^2
    • \left(k \left(\frac{ 2 q_1}{ 2d^2} \sin{ (\theta)}
    • \frac{ 2 q_2}{ d^2}\right)\right)^2}]
  • [\frac{ \hat E( P)}{ E( P)} = 2]
  • [\frac{ \hat E_x( P)}{ E_x( P)} = 2]
  • [\frac{ \hat E_y( P)}{ E_y( P)} = 2]

5)

How would you change [q_1] (keeping [q_2] and [q_3] fixed) in order to make the electric field at point [P] equal to zero?

• There is no change you can make to q_1 that will result in the electric field at point [P] being equal to zero.
  • The contribution tothe electric field at point [P] from [q_1] will be along the direction that connects the position of [q_1] to point [P]. The only way this contribution can cancel the contributions from [q_2] and [q_3] would be if the charges [q_2] and [q_3] were equal to each other so that the [x] and [y] components of the field produced by them would be equal. It is only in this case that it would be possible to adjust [q_1] to create a zero electric field at point [P].