Questions: Problem 2 A charge of (q1=4.0 mu mathrmC) is at the origin and a charge (q2=2.0 mu C) is at (x=-10 mathrm~m) along the (x)-axis. What is the magnitude and direction of the force on a charge, (q0=2.0 mu mathrmC), at (x=0 mathrm~m, y=1 mathrm~m), due to (q1) and (q2) ?

Solution Steps

• Calculate the force between \( q_0 \) and \( q_1 \) using Coulomb's Law.
• Calculate the force between \( q_0 \) and \( q_2 \) using Coulomb's Law.

• Distance between \( q_0 \) at \((0, 1)\) and \( q_1 \) at the origin \((0, 0)\) is \( r_{01} = \sqrt{(0-0)^2 + (1-0)^2} = 1 \, \text{m} \).
• Distance between \( q_0 \) at \((0, 1)\) and \( q_2 \) at \((-10, 0)\) is \( r_{02} = \sqrt{(-10-0)^2 + (0-1)^2} = \sqrt{101} \, \text{m} \).

• Use Coulomb's Law: \( F = k \frac{|q_1 q_2|}{r^2} \), where \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).
• Force between \( q_0 \) and \( q_1 \): \[ F_{01} = k \frac{|q_0 q_1|}{r_{01}^2} = 8.99 \times 10^9 \frac{(2.0 \times 10^{-6})(4.0 \times 10^{-6})}{1^2} \, \text{N} \]
• Force between \( q_0 \) and \( q_2 \): \[ F_{02} = k \frac{|q_0 q_2|}{r_{02}^2} = 8.99 \times 10^9 \frac{(2.0 \times 10^{-6})(2.0 \times 10^{-6})}{101} \, \text{N} \]

• \( F_{01} \) acts along the positive \( y \)-axis since \( q_0 \) is above \( q_1 \).
• \( F_{02} \) acts at an angle \(\theta\) with respect to the negative \( x \)-axis, where \(\theta = \tan^{-1}\left(\frac{1}{10}\right)\).

• \( F_{01} \) has components: \( F_{01x} = 0 \), \( F_{01y} = F_{01} \).
• \( F_{02} \) has components: \[ F_{02x} = F_{02} \cos(\theta), \quad F_{02y} = F_{02} \sin(\theta) \]

• Net force in \( x \)-direction: \( F_{net,x} = F_{02x} \).
• Net force in \( y \)-direction: \( F_{net,y} = F_{01y} + F_{02y} \).

• Magnitude: \[ F_{net} = \sqrt{F_{net,x}^2 + F_{net,y}^2} \]
• Direction: \[ \phi = \tan^{-1}\left(\frac{F_{net,y}}{F_{net,x}}\right) \]

Final Answer