Tutorial 3: Worked solutions

6a · Point-charge field

1. The source is at x = −a and the field point at x = a, so the displacement points along +x and has length 2a:

   R₂₁ = 2a e_x,    |R₂₁| = 2a

2. Coulomb’s field is the source charge times the radial unit vector divided by 4πε₀R²:

   E_Q1(Q₂) = Q₁4πε₀ |R₂₁|² e_x

   E_Q1(Q₂) = Q₁16πε₀ a² e_x

   The sign of Q₁ sets whether the field points +x or −x. The 1/(2a)² factor is easy to miss.

6b · Point-charge force

1. The electric force on a charge is its charge times the electric field at its position:

   F₁₂ = Q₂ E_Q1(Q₂)

2. Substitute the result from 6a:

   F₁₂ = Q₁ Q₂16πε₀ a² e_x

   Like signs give a force toward +x on Q₂; opposite signs reverse it. Its units are newtons because charge times field is C·N/C.

6c · Infinite-line fields

1. For λ₁, the displacement from its axis (0, a) to Q₂ (a, 0) is (a, −a). Its radius is √2 a and its radial unit vector is (e_x − e_y)/√2:

   E_λ1 = λ₁2πε₀ √2 a · e_x − e_y√2 = λ₁4πε₀ a(e_x − e_y)

2. For λ₂, the displacement is (a, a/2), with radius √5 a/2 and unit direction (2e_x + e_y)/√5:

   E_λ2 = λ₂πε₀ √5 a · 2e_x + e_y√5 = λ₂5πε₀ a(2e_x + e_y)

3. Add the x and y components separately:

   E_lines = 1πε₀ a[(λ₁4 + 2λ₂5)e_x + (−λ₁4 + λ₂5)e_y]

   Check the second line’s distance is a√5/2, not a√5; its source is only a/2 below the x-axis. Each field component has units N/C.

6d · Force from both lines

1. Both line fields act at Q₂, so their combined force is Q₂ times their vector sum:

   F_lines,2 = Q₂ E_lines

2. Therefore:

   F_lines,2 = Q₂πε₀ a[(λ₁4 + 2λ₂5)e_x + (−λ₁4 + λ₂5)e_y]

   A negative component indicates the corresponding force points along the negative axis; the force components are in newtons.

6e · Superposition of all forces

1. Electrostatic forces add linearly, so the total is the point-charge contribution plus the two-line contribution:

   F₂ = F₁₂ + F_lines,2

2. Collect the coefficients of each Cartesian direction:

   F₂ = [Q₁ Q₂16πε₀ a² + Q₂πε₀ a(λ₁4 + 2λ₂5)]e_x + Q₂πε₀ a(−λ₁4 + λ₂5)e_y

   Check: if both line densities vanish, this reduces to the Coulomb force in 6b.

6f · Balance the two components

1. Q₂ is nonzero, so the y-component vanishes when:

   −λ₁4 + λ₂5 = 0    ⇒    λ₂ = 5λ₁4

2. Insert this relation into the x-component, including the field from Q₁:

   Q₁16a² + λ₁4a + 25a · 5λ₁4 = 0

   Q₁16a² + 3λ₁4a = 0    ⇒    Q₁ = −12aλ₁

3. Thus the required condition is λ₂ = 5λ₁/4 and Q₁ = −12aλ₁ (equivalently λ₁ = −Q₁/(12a), λ₂ = −5Q₁/(48a)). These signs make the line forces cancel in y and balance Q₁’s x contribution.

7a · First known pair force

1. The displacement from Q₂ at (3a, 0) to Q₁ at (0, a) is (−3a, a), and the distance is √10 a:

   R₁₂ = a(−3e_x + e_y),    |R₁₂| = √10 a

2. Coulomb force on Q₁ from Q₂ is kQ₁Q₂ times the unit displacement divided by R². Here Q₁Q₂ = √10 Q²:

   F₁₂ = 14πε₀ · Q₁ Q₂|R₁₂|³ R₁₂ = 14πε₀ · √10 Q²10√10 a³ a(−3e_x + e_y)

   F₁₂ = Q²40πε₀ a²(−3e_x + e_y) = F₀(−3e_x + e_y)

   The displacement points from the source toward Q₁, so for like charges this force points up and to the left, away from Q₂.

7b · Equal and opposite force

1. Forces between a pair have equal magnitude and opposite direction:

   F₂₁ = −F₁₂

2. Therefore:

   F₂₁ = F₀(3e_x − e_y)

   The two pair forces sum to zero when applied to both charges together.

7c · Isolate the remaining force

1. The given resultant on Q₁ is F₁ = F₁₂ + F₁₃, so rearrange to isolate the force from Q₃:

   F₁₃ = F₁ − F₁₂

2. Subtract the components (the +e_y terms cancel):

   F₁₃ = F₀[−(3 + 5√5)e_x + e_y − (−3e_x + e_y)]

   F₁₃ = −5√5 F₀ e_x

   Its purely horizontal direction places Q₃ on the horizontal line through Q₁.

7d · Second remaining pair force

1. The given resultant on Q₂ is F₂ = F₂₁ + F₂₃. Since F₂₁ = −F₁₂:

   F₂₃ = F₂ − F₂₁ = F₂ + F₁₂

2. Combine components:

   F₂₃ = F₀[(13 − 3)e_x + (−11 + 1)e_y]

   F₂₃ = 10F₀(e_x − e_y)

   Its direction line is diagonal, with equal x and negative y components.

7e · Locate the third charge

1. F₁₃ has no y component, so the line joining Q₁ and Q₃ is horizontal. Since y₁ = a, this gives y₃ = a.

2. F₂₃ is parallel to (1, −1). The line through Q₂ = (3a, 0) in that direction has form (3a, 0) + t(1, −1). Its y-coordinate equals a when t = −a:

   (x₃, y₃) = (3a, 0) − a(1, −1) = (2a, a)

3. Result: Q₃ is at (2a, a). This position also gives Q₁Q₃ = 2a and Q₂Q₃ = √2 a.

7f · Find the third charge

1. The force magnitude between Q₁ and Q₃ is 5√5 F₀. Coulomb’s law at separation 2a gives:

   5√5 F₀ = 14πε₀ · √5 Q |Q₃|(2a)²

2. Insert F₀ = Q²/(40πε₀a²) and cancel common positive factors:

   5√5 Q²40πε₀ a² = √5 Q |Q₃|16πε₀ a²

   |Q₃| = 2Q

3. F₁₃ points from Q₃ toward Q₁, and Q₃ is to the right of Q₁; the force is repulsive. Thus Q₃ has the same sign as Q₁, and Q₃/Q = 2.

7g · Resultant on Q₃

1. The force on Q₃ from Q₁ is opposite the force of Q₃ on Q₁; the same reversal holds for the Q₂ pair:

   F₃₁ = −F₁₃ = 5√5 F₀ e_x,    F₃₂ = −F₂₃ = −10F₀ e_x + 10F₀ e_y

2. Now add the two forces acting on Q₃:

   F₃ = F₀[(5√5 − 10)e_x + 10e_y]

   Q₂, below and to the right of Q₃, pushes it up and to the left; Q₁, to its left, pushes it to the right. The two x parts nearly cancel, leaving a small positive x component and a positive y component. Numerically, 5√5 − 10 ≈ 1.18.