- #1
Jkalirai
- 4
- 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: q1 = +2.0 x 10-5 C, q2 = q3 = -3.0 x 10-5 C, r31 = r21 = 2 m
a) We start by finding the electric force between q3 to q1 and q2 to q1
FE31 = k * q1 * q3 / r312
FE31 = (9.0 x 109 Nm2/C2) * (+2.0 x 10-5 C) * (3.0 x 10-5 C) / (2 m)2
FE31 = 1.35 N
FE21 = k * q1 * q2 / r212
Since q2 = q3 and r31 = r21
FE21 = 1.35 N
Since the force of gravity is assumed to be negligible, we know that FE21 = FNETx and FE31 = FNETy
We can calculate FNET from our components,
FNET = √ [(1.35)2 + (1.35)2]
FNET = 1.9092 N
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 = 45o
Therefore, the net force on charge 1 is equal to 1.91 N [N 45o E].
b) To find ε on charge 1 we solve for the components of ε.
We know that q2 = q3 and r2 = r3
Therefore, ε2 = ε3
ε2 = ε3 = k * q / r2
ε = (9.0 x 109 Nm2/C2) * (3.0 x 10-5 C) / (2 m)2
ε = 6.75 x 104 N/C
εNET = √ [(6.75 x 104)2 + (6.75 x 104)2]
εNET = 95459.4155 N/C
Our angle will be,
Θ = tan-1 (6.75 x 104 N/C / 6.75 x 104 N/C)
Θ = 45o
Therefore, our net electric field acting on charge 1 is 9.55 x 104 N/C [N 45o 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.
a) We start by finding the electric force between q3 to q1 and q2 to q1
FE31 = k * q1 * q3 / r312
FE31 = (9.0 x 109 Nm2/C2) * (+2.0 x 10-5 C) * (3.0 x 10-5 C) / (2 m)2
FE31 = 1.35 N
FE21 = k * q1 * q2 / r212
Since q2 = q3 and r31 = r21
FE21 = 1.35 N
Since the force of gravity is assumed to be negligible, we know that FE21 = FNETx and FE31 = FNETy
We can calculate FNET from our components,
FNET = √ [(1.35)2 + (1.35)2]
FNET = 1.9092 N
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 = 45o
Therefore, the net force on charge 1 is equal to 1.91 N [N 45o E].
b) To find ε on charge 1 we solve for the components of ε.
We know that q2 = q3 and r2 = r3
Therefore, ε2 = ε3
ε2 = ε3 = k * q / r2
ε = (9.0 x 109 Nm2/C2) * (3.0 x 10-5 C) / (2 m)2
ε = 6.75 x 104 N/C
εNET = √ [(6.75 x 104)2 + (6.75 x 104)2]
εNET = 95459.4155 N/C
Our angle will be,
Θ = tan-1 (6.75 x 104 N/C / 6.75 x 104 N/C)
Θ = 45o
Therefore, our net electric field acting on charge 1 is 9.55 x 104 N/C [N 45o 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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