New capacitance when dielectric material covers bottom half

AI Thread Summary
The problem involves a parallel plate capacitor with a dielectric slab covering the bottom half, resulting in a new capacitance. The original capacitance is given by C0, and the new capacitance is derived from the series combination of two capacitors: one with a dielectric constant of K=2 for the lower half and one with K=1 for the upper half. The correct new capacitance is determined to be 4/3 C0, contrasting with an initial incorrect calculation of 3/2 C0. The area A of the plates remains unchanged, while the effective distance between the plates is halved due to the dielectric. The discussion emphasizes the importance of understanding series capacitance in the context of dielectrics.
Koh Eng Kiat
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Homework Statement



The plates of a parallel plate capacitor of capacitance C0 are
horizontal. Into the gap a slab of dielectric material with K=2 is placed,
filling the bottom half of the gap between the plates. What is the
resulting new capacitance?

Homework Equations



C = Q/V = κɛ0A/d

The Attempt at a Solution



Co = ɛ0A/d
Cf = kɛ0A/d = ((ɛ0)(0.5A)/d) + ((2ɛ0)(1/2a)/(d)) = (ɛ0A/2d) +(ɛ0A/d) = (ɛ0A/d) + (2ɛ0A/2d) = (3ɛ0A/2d)
Cf = 3/2 Co

The correct answer is 4/3 Co. Guys, can anyone help me out? I have no idea how to derive the correct answer.
 
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The new capacitance is a series combination of two capacitors, one with k=2 and the other with k=1(upper half). If the original capacitance was C, what values do you get for these new series capacitances?
 
Do you need to change A to 1/2A?
 
Koh Eng Kiat said:
Do you need to change A to 1/2A?
A remains the same. Distance between the plates is halved..
 
Thanks! I solved it already!
 
Hey guys, can you tell me how exactly can I get?
 
Amr Ayman said:
Hey guys, can you tell me how exactly can I get?
cnh1995 said:
A remains the same. Distance between the plates is halved..
cnh1995 said:
The new capacitance is a series combination of two capacitors, one with k=2 and the other with k=1(upper half). If the original capacitance was C, what values do you get for these new series capacitances?
 
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