A charge +Q and -2Q were placed on first and third of three very large parallel plates. The middle plate is earthed. Find the chargeon right extreem plate
Which plate is on the extreme right?
Are they equally spaced?
Right extreem means the 3rd plate to which -2Q chage was placed.
To find the charge on the right extreme plate, we need to consider the electric field created by the charged plates.
The first step is to determine the electric field created by the charges +Q and -2Q on the first and third plates, respectively. The electric field produced by a uniformly charged infinite parallel plate is given by the equation:
E = σ / (2ε₀),
where E is the electric field, σ is the surface charge density (charge per unit area) on the plate, and ε₀ is the permittivity of free space.
In this case, the charge +Q is on the first plate, and since the plates are very large, we can assume the charge is distributed uniformly on the plate's surface. Therefore, the surface charge density of the first plate is:
σ₁ = +Q / A,
where A is the area of the first plate.
Similarly, the charge -2Q on the third plate gives a surface charge density:
σ₃ = -2Q / A,
where A is the area of the third plate.
Now, let's consider the middle plate, which is earthed. When a conductor is connected to the Earth, its potential becomes zero. Therefore, the potential of the middle plate is zero.
Since the potential is constant throughout a conducting plate, the potential difference between the first and middle plates is the same as between the middle and third plates.
Let's assume the distance between the plates is d, and the electric field between the plates is E. The potential difference between the plates is then given by:
V = Ed.
Since the potential of the middle plate is zero, the potential difference between the first and third plates is:
V₁₃ = V - 0 = V.
The potential difference V₁₃ can also be expressed as the product of the electric field and the distance between the plates:
V₁₃ = E * d.
Now, since the potential difference between the first and third plates is the same, and the electric field is constant, we have:
E * d = E * d.
Using the equation E = σ / (2ε₀), we can rewrite this equation as:
(σ₁ / (2ε₀)) * d = (σ₃ / (2ε₀)) * d.
Plugging in the expressions for σ₁ and σ₃ obtained earlier, we get:
(+Q / (2ε₀A)) * d = (-2Q / (2ε₀A)) * d.
Simplifying this equation, we find:
Q = -2Q.
This equation tells us that the charge on the right extreme plate, Q, is equal to -2Q. In other words, the charge on the right extreme plate is -2 times the charge on the first plate.
Therefore, the charge on the right extreme plate is -Q.