Two point charges lie on the x axis. A charge of + 6.5 µC is at the origin, and a charge of -1.5 µC is at x = 10.0 cm.
What is the net electric field at x = -2.0 cm?
I found the electrical field for each charge using the equation E= k(q)/ r^2. My calculations were:
For the first charge- (8.99e9)(6.5e-6)/(.2^2) = 1460875 and (8.99e9)(1.5e-6)/(.2 + .1)^2 = 149833.33 for the second. I added the two to determine the net field- 1610708.33N/C. This answer was incorrect. So then I used -1.5e-6, recalculated and got net of 1311041.67; also incorrect. Can anyone tell me if I am I using the wrong equation or doing something else wrong?
You are doing something else wrong. The distance from charge 1 to the point where the field is desired is 0.02 m, not 0.2. Likewise for charge 2, it is 0.1 + 0.02m , not 0.1 + 0.2.
Using the corrected distances, we can calculate the electric fields due to each charge at x = -2.0 cm.
For the first charge (at the origin):
E₁ = k(q₁) / r₁² = (8.99e9)(6.5e-6) / (0.02²) = 1.44875e7 N/C
For the second charge (at x = 10.0 cm):
E₂ = k(q₂) / r₂² = (8.99e9)(-1.5e-6) / (0.12²) = -8.985e5 N/C
Since both fields are in the same direction (to the left, as they repel from the positive charge and attract to the negative one), we add the magnitudes to get the net electric field:
E_net = E₁ + E₂ = 1.44875e7 - 8.985e5 = 1.35891e7 N/C
To calculate the net electric field at x = -2.0 cm, you need to consider the individual electric fields due to each charge and then add them together.
The formula for the electric field due to a point charge is:
E = k(q) / r^2
Given:
q1 = +6.5 µC
q2 = -1.5 µC
x1 = 0.0 m
x2 = 0.10 m
Let's calculate the electric field due to charge 1:
r1 = distance from charge 1 to the point (-2.0 cm, 0.0 cm) = 0.02 m
E1 = (8.99 × 10^9 N m^2/C^2)(6.5 × 10^-6 C) / (0.02 m)^2 = 1185250 N/C (rounded to the nearest whole number)
Now, let's calculate the electric field due to charge 2:
r2 = distance from charge 2 to the point (-2.0 cm, 0.0 cm) = 0.12 m
E2 = (8.99 × 10^9 N m^2/C^2)(1.5 × 10^-6 C) / (0.12 m)^2 = 156250 N/C (rounded to the nearest whole number)
Finally, we can find the net electric field by adding the individual electric fields:
E_net = E1 + E2 = 1185250 N/C + 156250 N/C = 1341500 N/C
Therefore, the net electric field at x = -2.0 cm is 1,341,500 N/C.
To correctly calculate the net electric field at x = -2.0 cm, you need to use the correct distances in your equations. The distance from the first charge to the point where the field is desired is 0.02 m (since x = -2.0 cm = -0.02 m). The distance from the second charge to the point where the field is desired is the sum of the distance from the second charge to the origin (0.10 m) and the distance from the origin to x = -2.0 cm (0.02 m), which amounts to 0.12 m.
Let's recalculate the electric fields for each charge using the corrected distances:
For the first charge (q1 = +6.5 µC):
E1 = k * q1 / r1^2
= (8.99e9 N m^2/C^2) * (6.5e-6 C) / (0.02 m)^2
= 1476250 N/C (rounded)
For the second charge (q2 = -1.5 µC):
E2 = k * q2 / r2^2
= (8.99e9 N m^2/C^2) * (-1.5e-6 C) / (0.12 m)^2
= -1401042 N/C (rounded)
Now, we can find the net electric field at x = -2.0 cm (x = -0.02 m) by adding the electric fields due to each charge:
Net electric field = E1 + E2
= 1476250 N/C + (-1401042 N/C)
= 74909 N/C (rounded)
Therefore, the net electric field at x = -2.0 cm is approximately 74909 N/C.