Two charges are placed between the plates of a parallel plate capacitor. One charge is +q1 and the other is q2= +5.00x10^-6 C . The charge per unit area on each plate has a magnitude of sigma=1.30 x 10^-4 C/m^2. 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?

Hey, there is only one of me :(

You need to know the electric field E is constant between the plates and has value of E = sigma/eo = 1.3*10^-4/8.85*10^-12
= .147 *10^8

To find the distance r between the two charges, we can use Coulomb's Law to calculate the force between the charges and the electric field of the parallel plate capacitor.

The force between two charges is given by Coulomb's Law:

F = k * |q1| * |q2| / r^2

Where F is the force, k is the electrostatic constant (9 x 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.

The electric field of a parallel plate capacitor can be given as:

E = |sigma| / (2 * epsilon0)

Where E is the electric field, sigma is the charge per unit area, and epsilon0 is the permittivity of free space (8.85 x 10^-12 C^2/N m^2).

In this case, the force on q1 due to q2 is equal to the force on q1 due to the electric field. Therefore, we can equate these two forces:

k * |q1| * |q2| / r^2 = |q1| * E

Now, we can substitute the given values into this equation and solve for r:

k * |q1| * |q2| / r^2 = |q1| * |sigma| / (2 * epsilon0)

Canceling out |q1| from both sides, we get:

k * |q2| / r^2 = |sigma| / (2 * epsilon0)

Substituting the given values:

(9 x 10^9 N m^2/C^2) * (5.00 x 10^-6 C) / r^2 = (1.30 x 10^-4 C/m^2) / (2 * 8.85 x 10^-12 C^2/N m^2)

Simplifying the equation, we have:

(5.00 x 10^-6 C) / r^2 = 1.30 x 10^-4 C / (2 * 8.85 x 10^-12 C^2/N)

(5.00 x 10^-6 C) / r^2 = (1.30 x 10^-4 C) / (17.7 x 10^-12 C^2/N)

(5.00 x 10^-6 C) / r^2 = (7.34 x 10^-9 C^2/N)

To isolate r^2, we divide both sides by (5.00 x 10^-6 C):

1 / r^2 = (7.34 x 10^-9 C^2/N) / (5.00 x 10^-6 C)

1 / r^2 = 1.468 x 10^-3 N

Now, we can solve for r^2 by taking the reciprocal of both sides:

r^2 = 1 / (1.468 x 10^-3 N)

r^2 = 6.81 x 10^2 N/m^2

Finally, taking the square root of both sides, we find:

r = √(6.81 x 10^2 N/m^2)

r ≈ 26.1 meters

Therefore, the distance r between the two charges is approximately 26.1 meters.

To find the distance (r) between the two charges, we can use the equation for the electric field between two point charges:

F = k * (|q1*q2|) / r^2

where F is the electric force, k is the electrostatic constant (k ≈ 9 x 10^9 Nm^2/C^2), q1 and q2 are the charges, and r is the distance between them.

Given that the force on q1 due to q2 equals the force on q1 due to the electric field of the parallel plate capacitor, we can set up the equation:

F(q1 due to q2) = F(q1 due to capacitor electric field)

k * (|q1*q2|) / r^2 = q1 * E

where E is the electric field created by the parallel plate capacitor.

We can express the electric field E in terms of the surface charge density (sigma):

E = sigma / (2 * epsilon_0)

where epsilon_0 is the vacuum permittivity (approximately 8.85 x 10^-12 C^2 / Nm^2).

Substituting the value of E in the equation, we get:

k * (|q1*q2|) / r^2 = q1 * (sigma / (2 * epsilon_0))

Rearranging the equation to solve for r, we have:

r = sqrt((k * (|q1*q2|)) / (q1 * (sigma / (2 * epsilon_0))))

Plugging in the given values:

k ≈ 9 x 10^9 Nm^2/C^2
q1 = +q1 (unknown)
q2 = +5.00 x 10^-6 C
sigma = 1.30 x 10^-4 C/m^2
epsilon_0 ≈ 8.85 x 10^-12 C^2 / Nm^2

We can now substitute these values into the equation and solve for the distance r.

I will call q1 = q Now the force on q1 dues to the plates is:

F = q E = 0.147*10^8 q

The force on q due to the q2 is
F = 9*10^9 q (5*10^-6)/r^2

set those equal
q cancels
solve for r