particle 1 of charge q1 = 6.60×10-5 C and particle 2 of charge q2 = 3.30×10-4 C are fixed to an x axis, separated by a distance d = 0.100 m.
Calculate their net electric field E(x) as a function of x for the following positive and negative values of x, taking E to be positive when the vector E points to the right and negative when E points to the left.
What is E(-0.100)?
What is E(-0.040)?
What is E(0.080)?
What is E(0.130)?
To calculate the net electric field E(x) as a function of x, we need to consider the electric field due to particle 1, the electric field due to particle 2, and then sum up these two contributions.
The electric field due to a point charge q at a distance r from the charge is given by Coulomb's Law:
E = k * q / r^2
where k is the electrostatic constant (k = 9.0 x 10^9 N m^2/C^2). In this case, we have two fixed charges, so the net electric field is given by:
E_net(x) = E_1(x) + E_2(x)
where E_1(x) is the electric field at position x due to particle 1, and E_2(x) is the electric field at position x due to particle 2.
To calculate E_1(x) and E_2(x), we need to determine the distance between each particle and the position x. The distances are given by:
d_1 = x + 0.100
d_2 = d - x = 0.100 - x
Now we can substitute these distances into the formula for the electric field to calculate E_1(x) and E_2(x):
E_1(x) = k * q_1 / (x + 0.100)^2
E_2(x) = k * q_2 / (0.100 - x)^2
Let's calculate the net electric field at each of the given positions:
1. E(-0.100):
Substituting x = -0.100 into the above equations, we get:
E_1(-0.100) = k * 6.60 x 10^-5 / (0.100 - 0.100)^2 = k * 6.60 x 10^-5 / 0.100^2
E_2(-0.100) = k * 3.30 x 10^-4 / (0.100 + 0.100)^2 = k * 3.30 x 10^-4 / 0.200^2
Then, calculate the net electric field:
E_net(-0.100) = E_1(-0.100) + E_2(-0.100)
2. E(-0.040):
Substituting x = -0.040 into the above equations, we get:
E_1(-0.040) = k * 6.60 x 10^-5 / (0.100 - 0.040)^2 = k * 6.60 x 10^-5 / 0.060^2
E_2(-0.040) = k * 3.30 x 10^-4 / (0.100 + 0.040)^2 = k * 3.30 x 10^-4 / 0.140^2
Then, calculate the net electric field:
E_net(-0.040) = E_1(-0.040) + E_2(-0.040)
3. E(0.080):
Substituting x = 0.080 into the above equations, we get:
E_1(0.080) = k * 6.60 x 10^-5 / (0.100 + 0.080)^2 = k * 6.60 x 10^-5 / 0.180^2
E_2(0.080) = k * 3.30 x 10^-4 / (0.100 - 0.080)^2 = k * 3.30 x 10^-4 / 0.020^2
Then, calculate the net electric field:
E_net(0.080) = E_1(0.080) + E_2(0.080)
4. E(0.130):
Substituting x = 0.130 into the above equations, we get:
E_1(0.130) = k * 6.60 x 10^-5 / (0.100 + 0.130)^2 = k * 6.60 x 10^-5 / 0.230^2
E_2(0.130) = k * 3.30 x 10^-4 / (0.100 - 0.130)^2 = k * 3.30 x 10^-4 / (-0.030)^2
Then, calculate the net electric field:
E_net(0.130) = E_1(0.130) + E_2(0.130)
Now, you can substitute the values of k, q1, q2, and x, and use a calculator to compute the values of E_net(x) for each position.