IDNLearn.com: Your trusted source for finding accurate and reliable answers. Get the information you need from our community of experts who provide accurate and thorough answers to all your questions.
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 greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Thank you for choosing IDNLearn.com. We’re here to provide reliable answers, so please visit us again for more solutions.
