Investigators are exploring ways to treat milk for longer shelf life by using pulsed electric fields to destroy bacterial contamination. One system uses 8.0-cm-diameter circular plates separated by 0.95 cm. The space between the plates is filled with milk, which has the same dielectric constant as that of water. The plates are briefly charged to 30,000 V. What is the capacitance of the system? How much charge is on each plate when they are fully charged?

C = 3.77*10⁻¹⁰ F = 377 pF

Q = 1.13*10⁻⁵ C

Explanation:

Given

D = 8.0 cm = 0.08 m

d = 0.95 cm = 0.95*10⁻² m

k = 80.4 (dielectric constant of the milk)

V = 30000 V

C = ?

Q = ?

We can get the capacitance of the system applying the formula

C = k*ε₀*A / d

where

ε₀ = 8.854*10⁻¹² F/m

and A = π*D²/4 = π * (0.08 m) ²/4

⇒ A = 0.00502655 m²

then

C = (80.4) * (8.854*10⁻¹² F/m) * (0.00502655 m²) / (0.95*10⁻² m)

⇒ C = 3.77*10⁻¹⁰ F = 377 pF

Now, we use the following equation in order to obtain the charge on each plate when they are fully charged

Q = C*V

⇒ Q = (3.77*10⁻¹⁰ F) * (30000 V)

⇒ Q = 1.13*10⁻⁵ C