Practice Problems: The Electric Field Solutions

1. (easy) A small charge (q = 6.0 mC) is found in a uniform E-field (E = 2.9 N/C). Determine the force on the charge.
F = qE
F = (6x10^-3)(2.9) = 0.02 N

2. (easy) Find the electric field acting on a 2.0 C charge if an electrostatic force of 10500 N acts on the particle.
F = qE
10500 = (2.0)E
E = 5250 N/C

3. (easy) A dipole is set up with a charge magnitude of 2x10^-7 C for each charge (one is positive and the other is negative.) The distance between the charges is 0.15 m.  What are the magnitude and direction of the E-field at the midpoint of the dipole? (Assume the positive charge is on the left.) Also determine the force magnitude and direction for an electron at that position in the field.
The E-field from both charges will point to the right, thus the overall E-field is to the right.  The magnitude of the overall E-field is the addition of the two E-fields caused by the charges:
E = E₊ + E_- = kq/r² + kq/r² = kq(1/r² + 1/r²)
E = (9x10⁹)(2x10^-7)(1/(0.15/2)² + 1/(0.15/2)²)
E = 640000 N/C
The force on the electron is F=qE
F = (1.6x10^-19)(640000) = 1x10^-13N

4. (moderate) Two charges (q₁ and q₂) are located on the x axis on a coordinate system. They are both positive, but the second charge has twice the magnitude of the first. q₁ is at -0.5 m while q₂ is at +0.5 m. Determine the overall direction of the E-field at the various positions listed below:
A. At the origin
B. At x = 0 and y is negative
C. At x = -0.5 and y is positive
D. At x > 0 but x < +0.5 m and y = 0
E. At x > +0.5 m and y = 0
F. At x = +0.5 m and y > 0
Choose your answers from the following:
Choice 1: Along the +x-axis
Choice 2: Along the +y axis
Choice 3: Along the –x axis
Choice 4: Along the –y axis
Choice 5: Between 1 and 89 degrees from the +x axis
Choice 6: Between 91 and 179 degrees from the +x axis
Choice 7: Between 181 and 269 degrees from the +x axis
Choice 8: Between 271 and 359 degrees from the +x axis
The answers shown below are based on the convention that the field direction is in the same as the force direction on a small, positive test charge.
A. This position is equidistant to both charges. Charge q₁ produces an E-field along the +x axis while charge q₂ produces an E-field pointing along the –x axis. Since q₂ is larger, it produces a bigger E-field. The superposition of the fields shows an overall E-field along the –x axis. Choice 3.
B. This position is equidistant to both charges. Charge q₁ produces an E-field pointing into the 4^th quadrant while charge q₂ produces an E-filed pointing into the 3^rd quadrant. Since q₂ is larger, it produces a bigger E-field. The superposition of the fields shows an overall E-field in the 3^rd quadrant. Choice 7.
C. Charge q₁ produces an E-field pointing upward (+y) while charge q₂ produces an E-field pointing into the 2^nd quadrant. Depending upon the value of the y coordinate, the superpositioned E-field can be in any direction between 91 and 179 degrees. Choice 6.
D. Charge q₁ produces an E-field along the +x axis while charge q₂ produces an E-field pointing along the –x axis. Since q₂ is larger and closer, it produces a bigger E-field. The superposition of the fields shows an overall E-field along the –x axis. Choice 3.
E. Both charges produce an E-field along the +x axis. Thus the overall E-field is in that direction. Choice 1.
F. Charge q₂ produces an E-field pointing upward (+y) while charge q₁ produces an E-field pointing into the 1^st quadrant. Depending upon the value of the y coordinate, the superpositioned E-field can be in any direction between 1 and 89 degrees. Choice 5.

5. (moderate) Repeat the prevous question for points A,B,D and E, except assume that the first charge is positive and the second charge is negative.
A. This position is equidistant to both charges. Charge q₁ produces an E-field along the +x axis and charge q₂ produces an E-field that also points along the +x axis. Thus, the superposition of the fields shows an overall E-field along the +x axis. Choice 1.
B. This position is equidistant to both charges. Charge q₁ produces an E-field pointing into the 4^th quadrant while charge q₂ produces an E-field pointing into the 1^st quadrant. Since q₂ is larger, it produces a bigger E-field. The superposition of the fields shows an overall E-field in the 1st quadrant. Choice 5.
D. Charge q₁ produces an E-field along the +x axis and charge q₂ produces an E-field pointing along the +x axis. The superposition of the fields shows an overall E-field along the +x axis. Choice 1.
E. Charge q₁ produces an E-field along the +x axis and charge q₂ produces an E-field pointing along the -x axis. Since q₂ is larger and closer, it produces a bigger E-field. The superposition of the fields shows an overall E-field along the -x axis. Choice 3.

6. (moderate) Charge q₁ (positive) is located at position (0, 0.5m) and has a magnitude of 2.9x10^-6 C. Charge q₂ (same charge as q₁) is located at the origin. Assume that these charges are unable to move. A third charge (q₃ = +1.0x10^-9 C and m = 4.0x10^-25 kg) is located at (1.0 m, 0.25 m). Determine the force on and the acceleration of charge q₃ at this position, and describe the trajectory the third charge would take when released in the field caused by the other two charges.
The distance, r, from either q₁ or q₂ to q₃:
r² = 1² + (0.25)²
r = 1.03 m
The E-field from q₁ and q₂ can be calculated separately, then superpositioned:
E₁ = kq₁/r² = k(2.9x10^-6)/(1.03)² = 2.46x10⁴ N/C (pointing along the line that connects q₃ and q₁, away from q₃, into the 4^th quadrant, at 346⁰)
E₂ = kq₂/r² = k(2.9x10^-6)/(1.03)² = 2.46x10⁴ N/C (pointing along the line that connects q₂ and q₃, away from q₃, into the 1st quadrant, at 14⁰)
The y-components of the E-fields cancel out.
The x-components add together to point in the +x direction.
E_x = (2.46x10⁴cos346) + (2.46x10⁴cos14) = 4.8x10⁴ N/C
F = ma
qE = ma
1.0x10^-9(4.8x10⁴) = (4.0x10^-25)a
a = 1.2x10²⁰ m/s²
Once q₃ begins to move it will get further from both q₁ and q₂, but it will stay equidistant from both, ensuring that the net force is oriented at 0°. As it moves, the force (and the acceleration) will decrease. Thus, it will continue to speed up, but at a lower rate at time goes on.

7. (moderate) Two charged particles, one (with a charge of +2μC and a mass m) located on the origin of an axis system and a second (with a charge of +3μC and a mass of 2m) located at x = 1 m are exerting a force on each other. Determine the magnitude of the force and then describe the trajectory each particle will undergo, including their velocities and accelerations.
The instantaneous force magnitude they both exert on each other is determined by Coulomb's Law. Thus, for the initial positions:
F = kq₁q₂/r²  
F = (9x10⁹)(2x10^-6)(3x10^-6)/1² = 0.054 N
The particles will accelerate away from each other on a straight line. The smaller particle will move along the -x axis, while the larger particle will move along the +x axis. They will both speed up as time goes on, but the smaller particle will speed up faster because, with a lower mass, it will have a greater acceleration due to the common force. As they move apart the accelerations on each will decrease because the force will decrease. At some point the accelerations will be so small as to approach zero, and the particles will essentially stop speeding up and simply move away from each other at a constant speed.

8. (moderate) A charged particle (-3.0C with a mass of 0.0002 kg) is injected into an E-field with an initial speed of 2000 m/s along the +z axis. The E-field is uniform in this region (500 N/C), and directed in the +y direction. Determine the acceleration components for all three directions (x,y and z). Additionally, calculate the length of time needed for the particle to move 1x10⁸ m along the y direction (you decide whether the motion is along negative or positive y) and the distance moved along the other two axes over that time frame. Assume that the initial position of the particle is at the origin of the axis system.
The only acceleration will be in the -y direction as the E-field acts on the negative particle in a direction opposite to its own orientation. The accelerations in the x and z directions is zero.
|F| = |q|E
|F| = (3)(500) = 1500 N (in the -y direction)
F_y= ma_y
-1500 = 0.0002(a_y)
a_y = -7.5x10⁶ m/s²
Δy = v_oyt + ½a_yt²
-1x10⁸ = 0 + ½(-7.5x10⁶)t²
t = 5.2 s
Distance moved along z axis:
Δz = v_ozt + ½a_zt² = 2000(5.2) + 0 = 10400 m
Distance moved along x axis:
Δx = v_ozt + ½a_zt² = 0 + 0 = 0 m

9. (moderate) Based on the information shown in the sketch below, determine the trajectory of the positively charged particle as it enters into the E-fields shown.