Question

Find the two possible values x3,r and x3,ℓ for the position of charge 3.

Answer 1

Given:

q1 = q
q2 = 4q
q3 = q
d = 2.00 cm
xq1 = 0 cm
xq2 = 2.00 cm
q = 2.00 nC

Let us determine the forces:
F1on3 = K ⋅ |q1| ⋅ |q3| / r2 = K ⋅ |q| ⋅ |q| / (d1-3)2 = K ⋅ q2 / (d1-3)2
F2on3 = K ⋅ |q2| ⋅ |q3| / r2 = K ⋅ |4q| ⋅ |q| / (d3-2)2 = K ⋅ 4q2 / (d3-2)2

We're able to solve for the variables in both cases since the forces are supposed to equal each other in each case.

 

F1on3 = F2on3
K ⋅ q2 / x2 = K ⋅ 4q2 / (2.00 cm - x)2
1/x2 = 4/ [4.00 cm2 - (4.00 cm⋅x) + x2]
[4.00 cm2 - (4.00 cm ⋅ x) + x2] / x2 = 4
4.00 cm2 / x2 - (4.00 cm ⋅ x) / x2 + x2 / x2 - 4 = 0
4.00 cm2 / x2 - (4.00 cm) / x + 1 - 4 = 0
4.00 cm2 / x2 - (4.00 cm) /x - 3 = 0
3 = 4.00 cm2 / x2 - (4.00 cm) / x
3x2 = 4.00 cm - 4.00 cm ⋅ x
3x2 + 4.00 cm ⋅ x - 4.00 cm = 0

After we plug that into the quadratic equation, we get:
x = -2 & x = 2/3 
x3,r, x3,ℓ = 0.667,-2.00 cm, cm

Answer 2

The force exerted by the charges son charge q₃ is given by Coulomb's

law.

Responses:

The possible values $x_{3,r}$ and $x_{3,l}$ for the position of charge 3 are;

• $\underline{x_{3,r} = \dfrac{2}{3}}$

• $\underline{x_{3, l} = -2}$

Which method can be used to determine the position of charge 3?

The question is possibly part of an online question, where three charges, q₁, q₂, and q₃, are on the x-axis

q₁ = 1.0 nC

q₂ = 4.0 nC

q₃ = 1.0 nC

The distance between q₁ and q₂ = 2 cm

The force exerted by q₁ on q₃ = The force  exerted by q₂ on q₃

Required:

The possible values of $\mathbf{x_{3,r}}$ and $\mathbf{x_{3,l}}$ which are the position of charge q₃

Solution:

According to Coulomb's law, we have;

• $\mathbf{\dfrac{k \cdot q_1 \cdot q_3}{r_1^2}} = \dfrac{k \cdot q_2 \cdot q_3}{r_2^2}$

Where;

r₁ = x₃ = The distance of q₃ from q₁

r₂ = 2 - x₃ = The distance of q₃ from q₂

Therefore;

$\dfrac{k \times 1.0 \times 1.0}{x_3^2} = \mathbf{ \dfrac{k \times 4.0 \times 1.0}{(2 - x_3)^2}}$

Which gives;

(2 - x₃)² × 1.0 × 1.0 = x₃² × 4.0 × 1.0

(2 - x₃)² = 4·x₃²

4·x₃² - (2 - x₃)² = 0

3·x₃² + 4·x₃ - 4 = 0

3·x₃² - 2·x₃ + 6·x₃ - 4 = 0  where, 4·x = - 2·x₃ + 6·x₃

3·x₃² + 6·x₃ - 2·x₃ - 4 = 0

3·x₃·(x₃ + 2) - 2·(x₃ + 2) = 0

Factorizing gives;

(3·x₃ - 2)·(x₃ + 2) = 0

Which gives;

• $x_3 \ to \ the \ right \ of \ q_1, \ \underline{x_{3,r} = \dfrac{2}{3}} \ and \ x_3 \ to \ the \ left \ of \ q_1 \ \underline{x_{3, l} = -2}$

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