Physics, 08.02.2021 23:10 ivilkas23

Three point charges are arranged along the x axis. Charge q1 = -4.10 nC is located at x= 0.250 mand charge q2 = 2.20 nC is at x= -0.320 m . A positive point charge q3 is located at the origin. What must the value of q3 be for the net force on this point charge to have magnitude 3.10 μN ? (ans in nC)

What is the direction of the net force on q3? + or - direction

Where along the x axis can q3 be placed and the net force on it be zero, other than the trivial answers of x=+[infinity] and x=−[infinity]? (ans in m)

Answers: 2

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Answers

Answer from: angie249

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;

$F = \dfrac{k \cdot q_1 \cdot q_2}{r^2}$

The net force acting on the point charge, 'q₃', is given as follows;

$F_{NET} = \dfrac{k \cdot q_1 \cdot q_3}{x_1^2} + \dfrac{k \cdot q_2 \cdot q_3}{x_2^2}$

By substituting the given values, we have;

$F_{NET} = 3.10 \ \mu N = \dfrac{k \cdot (-4.10 \ nC) \cdot q_3}{(0.250 \ m) ^2} + \dfrac{k \cdot 2.20 \ nC \cdot q_3}{(-0.320 \ m)^2}$

$3.01 \ \mu N= q_3 \cdot \left ( \dfrac{k \cdot (-4.10 \ nC) }{(0.250 \ m) ^2} + \dfrac{k \cdot 2.20 \ nC }{(-0.320 \ m)^2} \right) = -\dfrac{14117 \ nC \cdot K}{320}$

$\therefore q_3 = 3.01 \ \mu N \times -\dfrac{320 \ m^2}{14117 \ nC \cdot K}$

K = 9 × 10⁹ N·m²·C⁻²

$\therefore q_3 = 3.01 \ \times 10^{-6} \ N \times -\dfrac{320 \ m^2}{14117 \times 10^{-9} \ C \times 9 \times 10^9 \ N \cdot m^2 \cdot C^{-2}}= -7.58 \ nC$

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;

$0 \ N= q_3 \cdot \left ( \dfrac{k \cdot (-4.10 \ nC) }{(0.57 - r_1 ) ^2} + \dfrac{k \cdot 2.20 \ nC }{(r_1)^2} \right)$

$\therefore \dfrac{k \cdot 4.10 \ nC }{(0.57 - r_1 ) ^2} = \dfrac{k \cdot 2.20 \ nC }{(r_1)^2}$

From which we have;

$\dfrac{4.10 \ nC }{(0.57 - r_1 ) ^2} = \dfrac{2.20 \ nC }{(r_1)^2}$

(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

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