between the plates of a parallel plate capacitor of capacity c, two parallel plates of the same material and area same as the plate of the original capacitor are placed. if the thick ness of these plates is equal to 1/5 th of the distance between the plates of the original capacitor then the capacity of the new capacitor is
Answers
when two plates are inserted then the new capacitance will be 5/3 times the original capacitance.
Explanation :
We know that the capacitance of a parallel plate capacitor is given by,
C = ε₀A/d
where A is the area of plate and d is the distance between the plates
but when a plate of same metal is inserted with thickness t, then the new capacitance is given by,
C₁ = ε₀A/(d - t)
given the thickness of the plates is 1/5th of the distance
=> t = d/5
=> thickness of 2 plates, t = 2d/5
putting this values in the above equation,
C₁ = ε₀A/(d - 2d/5)
=> C₁ = ε₀A/(3d/5)
=> C₁ = (5/3) x (ε₀A/d)
=> C₁ = 5C/3
Hence when two plates are inserted then the new capacitance will be 5/3 times the original capacitance.
Answer:
5/3 C
Explanation:
Answer given in the attachment..
pls..see it...
