EEI

Exercise 2: Worked solutions

Electrical Engineering I · Module 92315 · Exercise 2

Try each part before opening the answer. The steps build fields and forces from direction and distance, then check against symmetry, units, or a limiting case.

Take the solutions offline

Download clean PDFDownload LaTeX source (.tex)

Task 4: Electric force compared with gravity

Protons at A carry QA = +1 C and electrons at B carry QB = −1 C; they are 1 km apart. Find |FE|/|FG|. Use γ = 6.667 × 10−11 m3 kg−1 s−2, e = 1.602 × 10−19 C, mp = 1.672 × 10−27 kg and me = 9.1094 × 10−31 kg.

Connected topic: Coulomb’s law (§1.2) gives the electric force; compare it with the supplied gravitational law.

4: Magnitude ratio

Try first: convert each 1 C charge quantity into a number of elementary charges, then form both force magnitudes before dividing.

Attempted it? Open the worked steps

4 · Coulomb force versus gravity

  1. Gravity needs group masses, so first convert total charge to particle count. Each particle has charge magnitude e:

    N = |Q|e;   mA = |QA|mpe;   mB = |QB|mee

  2. Coulomb’s law gives electric magnitude; opposite signs make the force attractive:

    |FE| = k|QAQB|rAB2,   k = 14πε0

  3. Insert the group masses into the given gravity relation:

    |FG| = γmAmBrAB2 = γ|QAQB|mpmee2rAB2

  4. Divide the magnitudes; the charge quantities and separation cancel:

    |FE||FG| = k e2γmpme

  5. Substitute k = 8.9877 × 109 N m2/C2 and the given constants:

    |FE||FG| ≈ 2.27 × 1039

    Result: electric force is about 2.27 × 1039 times stronger in magnitude.

    For a feel of the sizes: |FE| = k(1 C)2/(1000 m)2 ≈ 8.99 × 103 N, roughly the weight of a 900 kg car, while the two particle groups weigh only mA ≈ 1.04 × 10−8 kg and mB ≈ 5.69 × 10−12 kg, so |FG| ≈ 3.96 × 10−36 N.

    Check: the ratio is dimensionless and independent of distance since both laws vary as 1/r2.

Task 5: Field from two charges on the y-axis

Place Q1 at (0, a) and Q2 at (0, −a). Find the field throughout the xy-plane and simplify it on the x-axis for the stated charge choices. With charges in coulombs and lengths in metres, field values are in N/C.

Connected topics: Electric field strength (§1.3) defines one source field; superposition (§1.4) adds them.

Resolve each contribution from its own source. P is arbitrary; its displacement from each charge sets direction and distance.

5a: Field anywhere in the plane

Try first: write the displacement from each source to P = (x, y), then add both vectors.

Attempted it? Open the worked steps

5a · General field by superposition

  1. Point each displacement from source to P by subtracting source coordinates from point coordinates:

    r1 = x ex + (y − a)ey;   r2 = x ex + (y + a)ey

  2. A point-charge vector field is kQr/r3; signed Q handles direction. Superpose the two source fields:

    E(P) = kQ1r1|r1|3 + kQ2r2|r2|3

  3. Take the lengths from the displacement components and substitute:

    E(x,y) = kQ1[x ex + (y − a)ey][x2 + (y − a)2]3/2 + kQ2[x ex + (y + a)ey][x2 + (y + a)2]3/2

  4. Result: valid except at (0, ±a), where an ideal point-charge field is singular.

    Check: a positive source points along its source-to-P displacement; a negative source reverses it.

5b: Equal charges, Q1 = Q2 = Q

Try first: set y = 0 and identify which components cancel by symmetry.

Attempted it? Open the worked steps

5b · Equal charges on the x-axis

  1. At P = (x, 0), the displacements have equal lengths and opposite vertical components:

    r1 = x ex − a ey;   r2 = x ex + a ey;   |r1| = |r2| = √x2 + a2

  2. With equal Q, vertical components cancel and horizontal components add:

    E(x,0) = kQ(r1 + r2)(x2 + a2)3/2 = 2kQxex(x2 + a2)3/2

  3. Result: for positive Q it points +x when x > 0 and reverses for x < 0; negative Q reverses these directions.

    Check: at x = 0 symmetry requires zero, which the expression gives.

5c: Opposite charges, Q1 = Q and Q2 = −Q

Try first: again set y = 0. Opposite signs change which components reinforce.

Attempted it? Open the worked steps

5c · Opposite charges on the x-axis

  1. The distances still match, but signed contributions use the difference of the displacement vectors:

    r1 − r2 = −2a ey

  2. Horizontal parts cancel; vertical contributions add downward:

    E(x,0) = kQ(r1 − r2)(x2 + a2)3/2 = −2kQaey(x2 + a2)3/2

  3. Result: for positive Q it points from upper positive to lower negative: −y. Negative Q reverses it.

    Check: at the midpoint both fields point down. Far along the axis, magnitude falls as 1/|x|3, as for a dipole.

Task 6: Charges at the corners of a square

Four nonzero point charges occupy corners of a square of side a. Q is at the origin, Q2 lies on the positive y-axis, and the surrounding permittivity is ε0. With SI values, force is measured in newtons and work in joules.

Connected topics: Coulomb’s law (§1.2) gives each force; superposition (§1.4) combines them. Work connects to electrostatic potential (§1.8).

The square is rotated in the coordinate plane. Sloping sides are a; the vertical diagonal Q to Q2 is √2a.

6a: Force on Q at the origin

Try first: find the source coordinates from the diagram, then resolve each force into x and y components.

Attempted it? Open the worked steps

6a · Vector sum at the origin

  1. The square’s 45° sides project equally on both axes. The other corners and their distances from Q are:

    Q1: (−a√2, a√2);   Q2: (0, √2a);   Q3: (a√2, a√2)

    Side charges are distance a; Q2 is distance √2a.

  2. The top charge contributes along −y for positive QQ2; its distance is √2a:

    F2 = −kQQ22a2ey

  3. The side-charge unit directions from each source toward the origin give:

    F1 = kQQ1√2a2(ex − ey);   F3 = −kQQ3√2a2(ex + ey)

  4. Add components and factor the scale, where k = 1/(4πε0):

    F = kQa2 [Q1 − Q3√2 ex − (Q1 + Q3√2 + Q22)ey]

  5. Result: this vector is the net force on Q; charge signs select attraction or repulsion.

    Check: each term has charge product over distance squared. If Q1 = Q3, symmetry removes the x-component.

6b: Condition for zero force

Try first: set both components from 6a to zero. The central charge Q is nonzero.

Attempted it? Open the worked steps

6b · Cancel both components

  1. Since Q ≠ 0, the shared factor kQ/a² cannot make the force zero. The x-component requires:

    Q1 − Q3√2 = 0   ⟹   Q1 = Q3

  2. The y-component must also vanish:

    Q1 + Q3√2 + Q22 = 0

  3. Use Q3 = Q1 and solve:

    Q2 = −√2(Q1 + Q3) = −2√2Q1

  4. Result: Q1 = Q3 and Q2 = −2√2Q1.

    Check: equal side charges cancel horizontally; the top charge must have the opposite sign so its vertical force balances theirs.

6c: Work to move Q2 to (0, a)

Try first: with Q1 = Q3 = 0, compare its initial and final distance from Q. Take work as external work for a quasistatic move.

Attempted it? Open the worked steps

6c · Change in electrostatic potential energy

  1. The diagram places Q2 at (0, √2a); the requested endpoint is (0, a). Only the interaction with Q remains. This starting position is read from the diagram.

  2. For two point charges, U = kQQ2/r. Quasistatic external work equals the potential-energy change:

    Wext = ΔU = kQQ2(1a − 1√2a)

  3. Use k = 1/(4πε0):

    Wext = QQ24πε0a(1 − 1√2)

  4. Result: the signed external work is positive for QQ2 > 0 and negative for QQ2 < 0.

    If the task is read as bringing Q2 in from far away (r → ∞) instead of from its corner, the 1/(√2a) term becomes 0 and Wext = QQ2/(4πε0a). Say which starting point you use.

    Check: the final separation is smaller. Like charges need positive external work to move closer; unlike charges release energy.

Personal worked-solution guide based on the supplied EEI Exercise 2 sheet, written in my own words; no exercise text or figures are reproduced. Not an official FAU publication; check course details against current material. Made by Thanate Amornjiraporn (Focus).