Two parallel plate capacitors are constructed as shown below. Each capacitor has

Question

Two parallel plate capacitors are constructed as shown below. Each capacitor has plates with area A that are separated by a distance d. Two slabs of dielectric materials (with dielectric constants k and 2k) are placed in between the plates of each capacitor. In the left capacitor, each slab has a thickness d/2 and covers the entire plate area. In the right capacitor, each slab has a thickness d. and covers half of the plate area. Calculate the total capacitance of each of these capacitors. Express your answers in terms of kappa, A, and d. If these two capacitors each have the same voltage V across them, then find the relative value of the charge between the two capacitors (in other words, find Q_left/Q_right). Which capacitor has more charge? If each capacitor has the same voltage V across it, what is the magnitude of the electric field inside the 2k dielectric? Do this for each capacitor. You answers will include V, d, and kappa.

Explanation / Answer

a)

for left part:

let,

C1=K*eo*A/(d/2)

C2=2K*eo*A/(d/2)

equivalent of C1 and C2 is, Cnet=C1+C2

Cnet=K*eo*A/(d/2)+2K*eo*A/(d/2)

Cnet=3*K*eo*A/(d/2)

Cnet=6*K*eo*A/d

for right part,

C1=K*eo*A/2d

C2=2K*eo*A/2d

equivalent of C1 and C2 is, 1/Cnet=1/C1+1/C2

1/Cnet=1/(K*eo*A/2d) + 1/(2K*eo*A/2d)

======. Cnet=(K*eo*A/3d)

b)

q_left/q_right=(Cnet*v)_left/(cnet*v)_right

q_left/q_right=(6*K*eo*A/d)/(K*eo*A/3d)

q_left/q_right=18

===>

q_left has more charge

c)

here,

left side,

E=q/(2K*eo*A)

for right side,

E=q/(2K*eo*(A/2))

E=q/(eo*A)

here, electric field is not same

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