I have these problems that I keep doing and keep missing. If someone could please help me with what I am doing wrong!
Two charges, -25 µC and +15 µC, are fixed in place and separated by 3.4 m.
(a) At what spot along a line through the charges is the net electric field zero? Locate this spot relative to the positive charge. (Hint: The spot does not necessarily lie between the two charges.)
(b) What would be the force on a charge of +16 µC placed at this spot?
I did this a bunch of ways. One was saying E=kq/r.
So I got E=k(-25x10^-6)/x^2 + k(15x10^-6)/(3.4-x)^2 I had no idea where to go from there
Another way was (25)(x)^2 = 15(x-3.4)^2 From here I got the crazy answer of 4.164132563.
2. Two charges are placed between the plates of a parallel plate capacitor. One charge is +q1 and the other is q2 = +2.50 µC. The charge per unit area on each plate has a magnitude of σ = 2.00 10-4 C/m2. The force on q1 due to q2 equals the force on q1 due to the electric field of the parallel plate capacitor. What is the distance r between the two charges?
I used E=sigma/Eo which was 2x10^-4/8.85x10^-12. This came to 2.2598870.06
Then I did F=qE and F=kq1q2/r^2.
I set this = to each other with the q’s cancelling out and got 100. Which is wrong.
1. one charge negative, one charge positive. You wont get a solution in between the charges, because the E is in the SAME direction, they are not in opposite direction, so they will never cancel.
So, the position of x is outside the charges. Let x be the distance from one, 3.4+x be the other distance.
For the first problem, you are on the right track with using the formula for electric field (E = kq/r^2), but you need to remember that electric field is a vector quantity. The electric fields produced by the two charges will have opposite directions since one is positive and the other is negative. Therefore, when finding the net electric field, you need to subtract the magnitude of one electric field from the other.
Here's how you can proceed:
1. Calculate the magnitude of the electric field contributed by each charge using the formula E = kq/r^2.
- For the -25 µC charge, E1 = k(-25x10^-6)/r^2.
- For the +15 µC charge, E2 = k(15x10^-6)/(3.4-r)^2.
2. Set up an equation to find the distance (r) where the net electric field is zero: E1 - E2 = 0.
3. Solve the equation to find the value of r.
For the second problem, you again started correctly by using the formula for electric field (E = σ/ε0). However, the formula for the force on a charge (F = qE) should be used separately to find the force on q1 due to q2 and the force on q1 due to the electric field.
Here's the step-by-step process:
1. Calculate the magnitude of the electric field using the formula E = σ/ε0, where σ is the charge per unit area (2x10^-4 C/m^2) and ε0 is the permittivity of free space (8.85x10^-12 C^2/(N·m^2)).
2. Calculate the force on q1 due to q2 using the formula F1-2 = k(q1)(q2)/r^2, where k is the electrostatic constant (9x10^9 N·m^2/C^2).
3. Set up an equation by equating the force on q1 due to q2 to the force on q1 due to the electric field: F1-2 = q1E.
4. Solve the equation to find the distance (r) between the two charges.
By following these steps correctly, you should be able to find the correct answers to both problems.