Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A 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
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
