Three Point Charges Problem - Detailed Solution and Interactive Quiz

Three Point Charges Problem

Question: Three point charges are placed in the xy-plane as follows:

• A charge of \( +2\,\mu\text{C} \) at \( (0,0) \) meters
• A charge of \( +2\,\mu\text{C} \) at \( (1,0) \) meters
• A charge of \( -1\,\mu\text{C} \) at \( \left(0.5,\, \frac{\sqrt{3}}{2}\right) \) meters

Find the magnitude of the net force acting on the \( -1\,\mu\text{C} \) charge. Use Coulomb's constant \( k = 9 \times 10^9\, \text{N·m}^2\text{C}^{-2} \).

(a) \( 0.018\,\text{N} \)

(b) \( 0.018\sqrt{3}\,\text{N} \)

(c) \( 0.036\,\text{N} \)

(d) \( 0.036\sqrt{3}\,\text{N} \)

Detailed Step-by-Step Explanation

Step 1: Identify the Charges and Their Positions
We have three charges:

• \( q_1 = +2\,\mu\text{C} \) at \( A(0,0) \)
• \( q_2 = +2\,\mu\text{C} \) at \( B(1,0) \)
• \( q_3 = -1\,\mu\text{C} \) at \( C\left(0.5,\,\frac{\sqrt{3}}{2}\right) \)

Step 2: Calculate the Force on \( q_3 \) due to \( q_1 \) and \( q_2 \)
Coulomb's law gives the magnitude of the force between two point charges:

\( F = \frac{k |q_1 q_2|}{r^2} \)

Force from \( q_1 \) on \( q_3 \) ( \( F_{13} \) ):
Distance between \( q_1 \) and \( q_3 \):

\( r_{13} = \sqrt{\left(0.5-0\right)^2 + \left(\frac{\sqrt{3}}{2}-0\right)^2} = \sqrt{0.25 + \frac{3}{4}} = \sqrt{1} = 1 \, \text{m} \)

So,

\( F_{13} = \frac{9 \times 10^9 \times |2 \times (-1)|\times10^{-12}}{1^2} = \frac{9 \times 10^9 \times 2 \times 10^{-12}}{1} = 0.018\,\text{N} \)

Force from \( q_2 \) on \( q_3 \) ( \( F_{23} \) ):
Distance between \( q_2 \) and \( q_3 \):

\( r_{23} = \sqrt{\left(0.5-1\right)^2 + \left(\frac{\sqrt{3}}{2}-0\right)^2} = \sqrt{(-0.5)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{0.25 + \frac{3}{4}} = \sqrt{1} = 1 \, \text{m} \)

So,

\( F_{23} = \frac{9 \times 10^9 \times |2 \times (-1)|\times10^{-12}}{1^2} = 0.018\,\text{N} \)

Step 3: Determine the Directions of the Forces
Both forces \( F_{13} \) and \( F_{23} \) are attractive (since \( q_3 \) is negative and the other charges are positive) and thus point from \( q_3 \) toward \( q_1 \) and \( q_2 \) respectively.

Step 4: Vector Addition of the Forces
Note the geometry:
The positions of \( q_1 \), \( q_2 \), and \( q_3 \) form an equilateral triangle of side 1 m.
Therefore, the forces \( \vec{F}_{13} \) and \( \vec{F}_{23} \) on \( q_3 \) have equal magnitude \( 0.018\,\text{N} \) and are directed along the lines joining \( q_3 \) to \( q_1 \) and \( q_3 \) to \( q_2 \).

The angle between these two force vectors is \( 60^\circ \) (since all angles in an equilateral triangle are \( 60^\circ \)).

The resultant magnitude \( F \) is given by the law of cosines for vector addition:

\( F = \sqrt{F_{13}^2 + F_{23}^2 + 2F_{13}F_{23}\cos60^\circ} \)

Since \( \cos60^\circ = 0.5 \) and \( F_{13} = F_{23} = 0.018\,\text{N} \):

\( F = \sqrt{(0.018)^2 + (0.018)^2 + 2(0.018)(0.018)(0.5)} \)

Simplify the expression:

\( F = \sqrt{2(0.018)^2 + (0.018)^2} = \sqrt{3(0.018)^2} = 0.018\sqrt{3}\,\text{N} \)

Final Answer: The magnitude of the net force acting on the \( -1\,\mu\text{C} \) charge is option (b).