- #1

Jkalirai

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- 1

- Homework Statement
- a) Find the net force on charge 1.

b) What is the net electric field acting on charge 1?

- Relevant Equations
- F[SUB]E[/SUB] = k * q[SUB]1[/SUB] * q[SUB]2[/SUB] / r[SUP]2[/SUP]

ε = k * q / r^2

We are given: q

a) We start by finding the electric force between q

F

F

F

F

Since q

F

Since the force of gravity is assumed to be negligible, we know that F

We can calculate F

F

F

The angle of our net force can be calculated as well,

tanΘ = (1.35 N / 1.35N)

Θ = tan-1 (1.35N / 1.35 N)

Θ = 45

Therefore, the net force on charge 1 is equal to 1.91 N [N 45

b) To find ε on charge 1 we solve for the components of ε.

We know that q

Therefore, ε

ε

ε = (9.0 x 10

ε = 6.75 x 10

ε

ε

Our angle will be,

Θ = tan-1 (6.75 x 104 N/C / 6.75 x 104 N/C)

Θ = 45

Therefore, our net electric field acting on charge 1 is 9.55 x 10

I think my answer for part a) seems correct but for part b), I wanted to make sure it was right. It does seems to make sense when I think about it in my head, obviously the electric fields that are having an effect on charge A would be from charge B and C and their components would give me the net electric field. I've seen various different answers for part b) so I thought maybe I might be missing something. Anyways, I appreciate any input and thanks for giving this a read.

_{1}= +2.0 x 10^{-5}C, q_{2}= q_{3}= -3.0 x 10^{-5}C, r_{31}= r_{21}= 2 ma) We start by finding the electric force between q

_{3}to q_{1}and q_{2}to q_{1}F

_{E31}= k * q_{1}* q_{3}/ r_{31}^{2}F

_{E31}= (9.0 x 10^{9}Nm^{2}/C^{2}) * (+2.0 x 10^{-5}C) * (3.0 x 10^{-5}C) / (2 m)^{2}F

_{E31}= 1.35 NF

_{E21}= k * q_{1}* q_{2}/ r_{21}^{2}Since q

_{2}= q_{3}and r_{31}= r_{21}F

_{E21}= 1.35 NSince the force of gravity is assumed to be negligible, we know that F

_{E21}= F_{NETx}and F_{E31}= F_{NETy}We can calculate F

_{NET}from our components,F

_{NET}= √ [(1.35)^{2}+ (1.35)^{2}]F

_{NET}= 1.9092 NThe angle of our net force can be calculated as well,

tanΘ = (1.35 N / 1.35N)

Θ = tan-1 (1.35N / 1.35 N)

Θ = 45

^{o}Therefore, the net force on charge 1 is equal to 1.91 N [N 45

^{o}E].b) To find ε on charge 1 we solve for the components of ε.

We know that q

_{2}= q_{3}and r_{2}= r_{3}Therefore, ε

_{2}= ε_{3}ε

_{2}= ε_{3}= k * q / r^{2}ε = (9.0 x 10

^{9}Nm^{2}/C^{2}) * (3.0 x 10^{-5}C) / (2 m)^{2}ε = 6.75 x 10

^{4}N/Cε

_{NET}= √ [(6.75 x 10^{4})^{2}+ (6.75 x 10^{4})^{2}]ε

_{NET}= 95459.4155 N/COur angle will be,

Θ = tan-1 (6.75 x 104 N/C / 6.75 x 104 N/C)

Θ = 45

^{o}Therefore, our net electric field acting on charge 1 is 9.55 x 10

^{4}N/C [N 45^{o}E].I think my answer for part a) seems correct but for part b), I wanted to make sure it was right. It does seems to make sense when I think about it in my head, obviously the electric fields that are having an effect on charge A would be from charge B and C and their components would give me the net electric field. I've seen various different answers for part b) so I thought maybe I might be missing something. Anyways, I appreciate any input and thanks for giving this a read.

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