At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
Answer:
1) q₃ is approximately -7.58 nC
2) The direction of the net force on 'q₃' is the negative '-' direction
3) For the net force on 'q₃' to be zero, 'q₃' can be placed at x = -0.079 m or x = -1.881 m
Explanation:
1) The force, 'F', between two charged spheres is given as follows;
[tex]F = \dfrac{k \cdot q_1 \cdot q_2}{r^2}[/tex]
The net force acting on the point charge, 'q₃', is given as follows;
[tex]F_{NET} = \dfrac{k \cdot q_1 \cdot q_3}{x_1^2} + \dfrac{k \cdot q_2 \cdot q_3}{x_2^2}[/tex]
By substituting the given values, we have;
[tex]F_{NET} = 3.10 \ \mu N = \dfrac{k \cdot (-4.10 \ nC) \cdot q_3}{(0.250 \ m) ^2} + \dfrac{k \cdot 2.20 \ nC \cdot q_3}{(-0.320 \ m)^2}[/tex]
[tex]3.01 \ \mu N= q_3 \cdot \left ( \dfrac{k \cdot (-4.10 \ nC) }{(0.250 \ m) ^2} + \dfrac{k \cdot 2.20 \ nC }{(-0.320 \ m)^2} \right) = -\dfrac{14117 \ nC \cdot K}{320}[/tex]
[tex]\therefore q_3 = 3.01 \ \mu N \times -\dfrac{320 \ m^2}{14117 \ nC \cdot K}[/tex]
K = 9 × 10⁹ N·m²·C⁻²
[tex]\therefore q_3 = 3.01 \ \times 10^{-6} \ N \times -\dfrac{320 \ m^2}{14117 \times 10^{-9} \ C \times 9 \times 10^9 \ N \cdot m^2 \cdot C^{-2}}= -7.58 \ nC[/tex]
q₃ ≈ -7.58 nC
2) Given that the negative charge, 'q₁' (-4.10 nC), is located at x = 0.250 m, which is on the positive, '+' side of the origin, it will repel the negatively charged 'q₃', to the '-' direction. q₃ will also be attracted to the '-' direction by the positively charged 'q₂' which is at -0.320 m on the negative side of the origin
The net force's direction on q₃ will be in the '-' direction
3) For zero net force, we have;
The distance between the given point charges, r = 0.250 - (-0.320)) = 0.57 m
Let 'r1' represent the distance between 'q1' and 'q3', therefore, the distance between 'q2' and 'q3' is 0.57 - r1
By substitution, we have;
[tex]0 \ N= q_3 \cdot \left ( \dfrac{k \cdot (-4.10 \ nC) }{(0.57 - r_1 ) ^2} + \dfrac{k \cdot 2.20 \ nC }{(r_1)^2} \right)[/tex]
[tex]\therefore \dfrac{k \cdot 4.10 \ nC }{(0.57 - r_1 ) ^2} = \dfrac{k \cdot 2.20 \ nC }{(r_1)^2}[/tex]
From which we have;
[tex]\dfrac{4.10 \ nC }{(0.57 - r_1 ) ^2} = \dfrac{2.20 \ nC }{(r_1)^2}[/tex]
(r₁)²×4.10 = 2.20 × (0.57 - r₁)²
From the above equation, we have;
95,000·r₁²+ 125,400·r₁-35739 = 0
Solving, using a graphing calculator we get;
r₁ ≈ 0.241 or r₁ ≈ -1.561
Where, 'r₁', is measured from 'q₂', therefore, we have;
r₁ = 0.241 + (-0.320) ≈ -0.079 or r₁ = -1.561 + (-0.320) ≈ -1.881
Therefore, the charge 'q₃' can be placed at x = -0.079 or x = -1.881 for the ne force on it to be zero
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
