This question is incomplete, the complete question is;
Consider a capacitor made of two rectangular metal plates of length L and width W, with a very small gap s between the plates. There is a charge +Q on one plate and a charge −Q on the other. Assume that the electric field is nearly uniform throughout the gap region and negligibly small outside.
Calculate the attractive force that one plate exerts on the other.
Remember that one of the plates doesn't exert a net force on itself. (Enter the magnitude. Use any variable or symbol stated above along with the following as necessary : ε0.)
Answer:
the attractive force F that one plate exerts on the other is Q² / 2ε0⊥ω
Explanation:
Given the information in the questioN;
Electric field by one plate at the pos of other plate is expressed as;
E₊ = r / 2ε0 = Q/2ε0A
SO force on the other plate
F = /QE₊ / = Q² / 2ε0A
F = Q² / 2ε0⊥ω
Therefore the attractive force that one plate exerts on the other is F = Q² / 2ε0⊥ω
The attractive force which one plate exert on the other is
[tex]F = \frac{Q^2}{2\epsilon LW}[/tex]
From the attached diagram, the charge on the rectangular plates are shown;
since we are told the metal plates is rectangular, the area of the plate is
[tex]A = L * W[/tex]
The electrical field on one plate at one point against the other will be
[tex]E_r = \frac{r}{2\epsilon} \\E_r = \frac{Q}{2\epsilon A}\\[/tex]
The force on the other plate can be calculated as
[tex]F = |QE_r| = \frac{Q^2}{2\epsilon A}\\ F = \frac{Q^2}{2\epsilon LW}[/tex]
The attractive force which one plate exert on the other is
[tex]F = \frac{Q^2}{2\epsilon LW}[/tex]
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