A test charge of +5.0*10^-6 experiences a force of 2.0*10^3N(s) when placed at the midpoint of two oppositely charged parallel plates. Assuming that the plates are electrically isolated and have a distance separation of 8.0mm. What will be the force experienced by a different charge of -2.0*10^-6, located 2.0mm from the negative plate?
Well, let's calculate that using some electrical humor, shall we? Remember, electricity can be quite a shocking subject, so don't get too charged up!
Given:
Test charge (q₁) = +5.0*10^-6 C
Force (F₁) = 2.0*10^3 N
Plate separation (d) = 8.0 mm = 0.008 m
Now, using the formula for electric force between point charges: F = (k * q₁ * q₂) / r²
Here's where it gets interesting - the force experienced by a different charge is proportional to the inverse square of the distance between them. That's a mouthful, isn't it?
So, let's find the new force (F₂) experienced by our -2.0*10^-6 C charge, which is 2.0 mm = 0.002 m away from the negative plate.
First, let's calculate the ratio of distances: β = (0.008 m) / (0.002 m) = 4
Since the force is inversely proportional to the square of the distance, we square this ratio: β² = (4)² = 16
Now, let's find the new force (F₂): F₂ = (F₁) / β² = (2.0*10^3 N) / (16) = 125 N
So, applying some electrical humor to the situation, the force experienced by the -2.0*10^-6 C charge located 2.0 mm from the negative plate is quite an "attractive" force of 125 N! It seems like there's some positive charges being pulled towards those negative plates. Just remember to stay positive in life, even when dealing with electrical forces!
To find the force experienced by the second charge, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is given by the equation:
F = k * (|q1| * |q2|) / r^2
Where:
F is the force between the charges
k is the electrostatic constant (k = 8.99 * 10^9 N m^2/C^2)
|q1| and |q2| are the magnitudes of the charges
r is the distance between the charges.
In this case:
|q1| = 5.0 * 10^-6 C
|q2| = 2.0 * 10^-6 C
r = 2.0 mm = 2.0 * 10^-3 m
Let's calculate the force using Coulomb's Law:
F = (8.99 * 10^9 N m^2/C^2) * (|5.0 * 10^-6 C| * |2.0 * 10^-6 C|) / (2.0 * 10^-3 m)^2
F = (8.99 * 10^9 N m^2/C^2) * (10^-5 C^2) / (4.0 * 10^-6 m^2)
F = (8.99 * 10^9 * 10^-5 C^2) / (4.0 * 10^-6 m^2)
F = 8.99 * 10^4 N
Therefore, the force experienced by the second charge is 8.99 * 10^4 N.
To find the force experienced by a different charge, we can use Coulomb's Law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Here's how you can calculate the force experienced by the different charge:
1. Calculate the electric field between the plates:
The electric field (E) between the parallel plates can be found by dividing the force experienced by the test charge by the magnitude of the charge itself. So, using the equation E = F/q, where F is the force and q is the charge, we have:
E = 2000 N / (5.0 * 10^(-6) C)
E = 4.0 * 10^8 N/C
2. Calculate the force on the different charge:
To find the force on the different charge, we first need to determine the electric field intensity at its position. The electric field intensity at a point due to the presence of a charged plate is given by the equation E = σ/(2ε₀), where σ is the charge density on the plate (charge per unit area) and ε₀ is the permittivity of free space. Since the plates are oppositely charged, the electric field intensity due to the positive plate will have the same magnitude as the electric field intensity due to the negative plate.
a. Electric field intensity due to the positive plate:
The charge density on the positive plate is given by σ = q/A, where q is the magnitude of the charge on the positive plate and A is the area of the plate. Since the plates are oppositely charged, we can assume the same magnitude of charge on both plates.
q = 5.0 * 10^(-6) C
A = 2.0 * 10^(-3) m * 8.0 * 10^(-3) m
σ = (5.0 * 10^(-6) C ) / (2.0 * 10^(-3) m * 8.0 * 10^(-3) m)
b. Electric field intensity due to the negative plate:
Since the negative plate has a charge of -2.0 * 10^(-6) C, we can use the same formula and calculate the charge density on the negative plate:
q = -2.0 * 10^(-6) C
σ = (-2.0 * 10^(-6) C ) / (2.0 * 10^(-3) m * 8.0 * 10^(-3) m)
c. Electric field intensity at the position of the different charge:
Since the electric field is a vector quantity, we need to consider the electric field intensities due to both plates with their appropriate directions. In this case, the positive plate's electric field points towards the negative plate, while the negative plate's electric field points away from it. Therefore, the net electric field intensity at the position of the different charge is the difference between the electric fields due to the positive and negative plates.
E_net = E_+ - E_- = σ/(2ε₀) - σ/(2ε₀)
3. Calculate the force on the different charge:
The force experienced by a charge in an electric field is given by the equation F = q * E. In this case, the different charge is -2.0 * 10^(-6) C, and the electric field intensity at its position is given by E_net.
F = (-2.0 * 10^(-6) C) * E_net
By substituting the appropriate values into the equation, you can find the force experienced by the different charge located 2.0mm from the negative plate.