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Opus in profectus

Electric Potential

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Practice

practice problem 1

A charge of -1.0 μC is located on the y-axis 1.0 m from the origin at the coordinates (0,1) while a second charge of +1.0 μC is located on the x-axis 1.0 m from the origin at the coordinates (1,0). Determine the value of the following quantities at the origin…
  1. the magnitude of the electric field,
  2. the direction of the electric field,
  3. the electric potential (assuming the potential is zero at infinite distance), and
  4. the energy needed to bring a +1.0 μC charge to this position from infinitely far away.

solution

  1. Since the charges are identical in magnitude and equally far from the origin, we can do one computation for both charges.
     
    E =  kq
    r2
    E =  (9.0  × 109 Nm2/C2)(1.0  × 10−6 C)
    (1.0 m)2
    E = 9,000 N/C  
     

    Electric field lines come out of positive charges and go into negative charges. At the origin, this results in an electric field that points "left" (away from the positive change) and "up" (toward the negative charge). These two vectors form the legs of a 45°–45°–90° triangle whose sides are in the ratio 1:1:√2.

    ΣE = √2 × 9000 N/C = 12,700 N/C

  2. Moving "up" and to the "left" in equal amounts results in a 135° standard angle.
  3. Once again, since the charges are identical in magnitude and equally far from the origin, we only need to compute one number.
     
    V =  kq
    r
    V =  (9.0  × 109 Nm2/C2)(1.0  × 10−6 C)
    (1.0 m)
    V = 9,000 V  
     

    Electric potential is a scalar quantity. It doesn't have direction, but it does have sign. The positive charge contributes a positive potential and the negative charge contributes a negative potential. Add them up and watch them cancel.

    ΣV = 9000 V − 9000 V = 0 V

  4. The electric potential at a point in space is defined as the work per unit charge required to move a test charge to that location from infinitely far away.
     
    ΔV =  ΔUE
    q

    Algebra shows that work is charge times potential difference. Since the potential at the origin is zero, no work is required to move a charge to this point.

    ΔUE = qΔV
    ΔUE = (1.0 × 10−6 C)(0 V)
    ΔUE = 0 J

practice problem 2

A proton (mass m, charge +e) and an alpha particle (mass 4m, charge +2e) approach one another with the same initial speed v from an initially large distance. How close will these two particles get to one another before turning around?

solution

The kinetic energy of the moving particles is completely transformed into electric potential energy at the point of closest approach.
 
Ue  =  K  
 
k(e)(2e)  =  1  (m)v2 +  1  (4m)v2
r 2 2

Finish the algebra.

r =  4ke2
5mv2

practice problem 3

sketch-v.pdf
The diagram on the accompanying pdf file shows the location and charge of four identical small spheres. Find the electric potential at the five points indicated with open circles. Use these results and symmetry to find the potential at as many points as possible without additional calculation. Write your results on or near the points. Sketch at least 4 equipotential lines. Pick round values seperated by a uniform interval. At least one of the lines should be disconnected.

solution

Use the equation for the electric potential from a set of point charges.

V = k ∑ q
r

Since each charge is the same size, we can factor it out.

V = kq ∑  1
r

In order to save screen real estate, let's compute the product of the constants once…

kq = (9 × 109 Nm2/C2)(1 × 10−6 C) = (9000 Nm2/C)

…and the sum of the distances to the four charges five times…

∑  1  = 
1  +  1  +  1  +  1
r1 √8 m √8 m √8 m √8 m
∑  1  = 1.41421…m−1
r1
 
∑  1  = 
1  +  1  +  1  +  1
r2 √8 m √8 m √40 m √40 m
∑  1  = 1.02333…m−1
r2
 
∑  1  = 
1  +  1  +  1  +  1
r3 √20 m √20 m √68 m √68 m
∑  1  = 0.68974…m−1
r3
 
∑  1  = 
1  +  1  +  1  +  1
r4 2 m 2 m √20 m √20 m
∑  1  = 1.44721…m−1
r4
 
∑  1  = 
1  +  1  +  1  +  1
r5 2 m √20 m 6 m √52 m
∑  1  = 1.02894…m−1
r5

Apply it at each of the five locations, summing up the contributions of the four point charges.

V1 = (9000 Nm2/C)(1.41421…m−1)
V1 = 12,700 V
 
V2 = (9000 Nm2/C)(1.02333…m−1)
V2 = 9,210 V
 
V3 = (9000 Nm2/C)(0.68974…m−1)
V3 = 6,210 V
 
V4 = (9000 Nm2/C)(1.44721…m−1)
V4 = 13,000 V
 
V5 = (9000 Nm2/C)(1.23605…m−1)
V5 = 9,260 V

Record the numbers at as many symmetric locations as possible.

Sketch in the equipotentials.

practice problem 4

Write something completely different.

solution

Answer it.