Practice Problems: Electric Fields

1. (easy) What is the magnitude of a point charge whose E-field at a distance of 25 cm is 3.4 N/C?
E= kq/r²
3.4 = (9x10⁹)q/(0.25)²
q = 2.4x10^-11C

2. (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

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 direction 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 previous question for points A,B,D and E, except assume that the first charge is negative and the second charge is positive.
A. This position is equidistant to both charges. Charge q₁ produces an E-field along the -x axis and charge q₂ produces an E-filed that also points along the –x axis. Since q₂ is larger, it produces a bigger E-field. Thus, 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 2^nd 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.
D. Charge q₁ produces an E-field along the -x axis and charge q₂ produces an E-filed 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. Charge q₁ produces an E-field along the -x axis and charge q₂ produces an E-filed 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 1.

6. (moderate) Four equal charges are located on the corners of a square as shown below.  What is the magnitude of the E-field at the center position of the square?

The E-field at the center is the superposition of the E-fields from all 4 charges.  Since the E-field magnitude for all the charges is the same (E=kq/r²) with r being the distance from any one charge to the center, and the direction of each E-field is AWAY from the individual charge, the overall E-field is ZERO.

7. (hard) Find the E-field (both magnitude and direction) at the center of the square charge distribution shown below.  Note that the charges are NOT equal.  Assume that the sides of the square have a length L.


To solve this problem you need to superposition the E-fields of all four charges.  The distance (r) to the center for any charge is the same (L/√2). Use the standard coordinate system to measure the angles below.
The lower left charge produces an E-field pointing at 225° with a magnitude of
E= kq/r² = 2kq/L²
The upper left charge produces an E-field pointing at 315° with a magnitude of
E = kq/r² = 2kq/L²
The lower right charge produces an E-field pointing at 135° with a magnitude of
E = k2q/r² = 4kq/L²
The upper right charge produces an E-field pointing at 45° with a magnitude of
E= k2q/r² = 4kq/L²
The x-components of this superposition cancel out.  The y-components add together in the following manner:
E = [(2kq/L²)(sin 225) + (2kq/L²)(sin 315) + (4kq/L²)(sin 135) + (4kq/L²)(sin 45)]
E= (2√2)kq/L² (in the +y direction)