solve 2 and 3 A point charge q1 = -2.1 μC is located at the origin of a co-ordinate system. Another point charge q2 = 6.8 μC is located along the x-axis at a distance x2 = 7 cm from q1.1)What is F12,x, the value of the x-component of the force that q1 exerts on q2? -26.2285N 2)Charge q2 is now displaced a distance y2 = 3 cm in the positive y-direction. What is the new value for the x-component of the force that q1 exerts on q2?N 3)A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 7.4 N and points away from q1 and q3. What is the value (sign and magnitude) of the charge q3?μC y y42)y2

A point charge q1 = -2.1 μC is located at the origin of a co-ordinate system. Another point charge q2 = 6.8 μC is located along the x-axis at a distance x2 = 7 cm from q1. 1) What is F12,x, the value of the x-component of the force that q1 exerts on q2? -26.2285N

A point charge q1 = -2.1 μC is located at the origin of a co-ordinate system. Another point charge q2 = 6.8 μC is located along the x-axis at a distance x2 = 7 cm from q1. 1) What is F12,x, the value of the x-component of the force that q1 exerts on q2? -26.2285N

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Question

solve 2 and 3

A point charge q₁ = -2.1 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 6.8 μC is located along the x-axis at a distance x₂ = 7 cm from q₁.

What is F_12,x, the value of the x-component of the force that q₁ exerts on q₂?

-26.2285N

Charge q₂ is now displaced a distance y₂ = 3 cm in the positive y-direction. What is the new value for the x-component of the force that q₁ exerts on q₂?

A third point charge q₃ is now positioned halfway between q₁ and q₂. The net force on q₂ now has a magnitude of F_2,net = 7.4 N and points away from q₁ and q₃. What is the value (sign and magnitude) of the charge q₃?

μC

Expert Solution

Step 1

Here we need to solve only for part 2 and 3 as mentioned in the question.

$c h a r g e q_{1} = - 2.1 μ C l o c a t e d a t t h e o r i g i n (0, 0) c h a r g e q_{2} = 6.8 μ C l o c a t e d a t t h e p o i n t (7, 3) \vec{r} = 7 i + 3 j {|\vec{r}|}^{2} = 7^{2} + 3^{2} = 49 + 9 = 58 |\vec{r}| = \sqrt{58} F_{12} = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r^{2}} = 9 \times 10^{9} \times \frac{- 2.1 \times 6.8 \times 10^{- 6} \times 10^{- 6}}{{(\sqrt{58} \times 10^{- 2})}^{2}} = - 2.216 \times 10^{- 3} \times 10^{4} = - 22.16 N \tan θ = \frac{3}{7} θ = 23.198 ° F_{x} = F \cos θ = - 22.16 \cos 23.2 = - 20.36 N T h i s f o r c e w i l l a c t a l o n g t h e x - a x i s A n s w e r : f o r c e a l o n g x - d i r e c t i o n = - 20.36 N (n e g a t i v e s i g n i m p l i e s a n a t t r a c t i v e f o r c e)$

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