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Opus in profectus

# Electric Potential

## 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)
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 N m2/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.

E = √2 × 9,000 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 N m2/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.

V = 9,000 V − 9,000 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.

 ∆V = ∆UE 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.

 ∆UE = q∆V∆UE = (1.0 × 10−6 C)(0 V)∆UE = 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 below 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 N m2/C2)(1 × 10−6 C) = (9,000 N m2/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 = (9,000 N m2/C)(1.41421…m−1) V1 = 12,700 V V2 = (9,000 N m2/C)(1.02333…m−1) V2 = 9,210 V V3 = (9,000 N m2/C)(0.68974…m−1) V3 = 6,210 V V4 = (9,000 N m2/C)(1.44721…m−1) V4 = 13,000 V V5 = (9,000 N m2/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

Fission is the splitting of a heavy atomic nucleus into two roughly equal halves accompanied by the release of a large amount of energy. An atomic nucleus can be modeled as a sphere whose charge is distributed uniformly across its entire volume. Determine the energy released when a heavy nucleus undergoes nuclear fission using electrostatic principles.
1. Derive an equation for the electrostatic energy needed to assemble a charged sphere from an infinite swarm of infinitesimal charges located infinitely far away. (In other words, use calculus.) Let R be the sphere's radius, Q be its total charge, V be its volume, and ρ be its charge density.
2. Express the total energy of two half-sized spheres in terms of the energy of one whole sphere. Half-sized spheres have half the volume and half the charge of a whole sphere (because charge density is assumed to be constant).
3. Calculate the energy released when a nucleus of uranium 235 (the isotope responsible for powering some nuclear reactors and nuclear weapons) splits into two identical daughter nuclei. Give your final answer in the preferred unit for nuclear reactions, the megaelectronvolt. (A nucleus of 23592U has a radius of 5.8337 fm.)

#### solution

1. One way to make a big sphere to add layers to an already existing smaller sphere. Calculus allows us to start with an initial sphere with zero radius (r0 = 0), add layers to it of infinitesimal thickness (dr), and end up with a sphere with nonzero radius (r = R) by repeating the process an infinite number of times (). This calculus thing is pretty amazing.

The electrostatic potential energy of two point charges is given by…

 U = kq1q2 r

where…

 U = electric potential energy k = the electrostatic constant q1 = one point charge q1 = another point charge r = the separation between charges

In our sphere built up layer by layer, the first charge is a solid sphere with uniform charge density.

 q1 = ρ 4 πr3 3

The second charge is a thin spherical shell with the same charge density.

q2 = ρ(4πr2dr)

These two charges are effectively separated by the radius of the solid sphere. The energy equation then becomes a mess…

 R⌠⌡0 U = k (ρ 4 πr3) ρ(4πr2dr) 1 3 r

begging to be simplified…

 R⌠⌡0 U = 16π2ρ2k r4dr 3

and solved.

 U = 16π2ρ2kR5 15

We should now replace charge density with a more useful expression.

 ρ = Q = Q = 3Q V 43πR3 4πR3

Actually, it's the square of charge density we should eliminate.

 ρ2 = 9Q2 16π2R6

So let's do it.

 U = 16π2kR5 9Q2 15 16π2R6

Yay algebra!

 U = 3kQ2 5R
2. A half sized sphere has half the charge and half the volume but not half the radius. It's the cube root of a half the radius.

 V = 4 πr3 3
⇒
 V = 4 π ⎛⎜⎝ r ⎞3⎟⎠ 2 3 ∛2

A half sphere then has energy equal to…

 U½ = 3k(Q/2)2 = ∛2 U 5(R/∛2) 4

and two of them have energy equal to…

 2U½ = ∛2 U 2

I think it's more interesting to express the weird fraction as a decimal.

 2U½ = ∛2 = 0.62996… U 2

Splitting a charged sphere in half reduces potential energy to 63% — or results in a loss of 37%, if you prefer. (This assumes the two spheres are infinitely far away from each other, so their interaction adds no additional potential energy.)

3. Here's how I'd like to approach this problem. Start by determining the electric potential energy of a 23592U nucleus using the equation derived in part a.

 U = 3kQ2 5R
 U = 3(8.99 × 109 N m2/C2)(92 × 1.60 × 10−19 C)2 5(5.8337 × 10−15 m)
 U = 2.00 × 10−10 J

Convert that into megaelectronvolts by dividing by the elementary charge (to get it into electronvolts) and also by a million (since the prefix mega means a million).

 U = 2.00 × 10−10 J (1.60 × 10−19 C)(1,000,000)
 U = 1252 MeV

Then take 37% of that.

U = 0.370(1252 MeV) = 463 MeV

This is about twice what I expected, but I'll have to figure out the discrepancy some other time.