Question

The figure below shows a couple of capacitors connected in series. The central section can be moved along the vertical axis. What is the net capacitance of this system? L = 1 cm, d = 2 cm, and the surface area of each plate is 4 cm2. Enter answer to 3 significant figures and input numbers like 2000 as 2E3. d L

Answer 1

We know that capacitance of a capacitor is given by:

C = e0*A/D

e0 = 8.854*10^-12

A = Area of plates = 4 cm^2 = 4*10^-4 m^2

D = distance between plates

Now suppose at any given central section is 'd1' distance above the lower plate, and 'd2' distance below the upper plate. then

Capacitance of capacitor formed by lower plate and central section will be:

C1 = e0*A/d1

Capacitance of capacitor formed by upper plate and central section will be:

C2 = e0*A/d2

Now equivalent capacitance, when two capacitors are connected in series will be:

1/Ceq = 1/C1 + 1/C2

1/Ceq = 1/(e0*A/d1) + 1/(e0*A/d2) = d1/(e0*A) + d2/(e0*A)

1/Ceq = (d1 + d2)/(e0*A)

Ceq = e0*A/(d1 + d2)

Now from given situation:

d1 + d2 = d - L = 2 - 1 = 1 cm = 0.01 m

So,

Ceq = 8.854*10^-12*4*10^-4/0.01

Ceq = 3.54*10^-13 F = 3.54E-13 F

Let me know if you've any query.