Two parallel plates that are initially uncharged are separated by 1.7 mm, have only air between them, and each have surface areas of 16 cm 2. how much charge must be transferred from one plate to the other if 1.9 j of energy are to be stored in the plates? (ε 0 = 8.85 × 10 -12 c 2/n ∙ m 2)

Explanation:

The capacitor of a parallel-plate capacitor is given by:

where

A is the area of each plate

d is the separation between the plates

is the vacuum permittivity

The energy stored in a capacitor instead is given by

where

Q is the charge stored in each plate

Substituting the expression we found for C inside the last formula,

And re-arranging it

Now if we substitute

We find the charge stored on the capacitor:

117.83° F

Explanation:

Using Newton's Law of Cooling which can be expressed as:

The differential equation can be computed as:

where;

At the initial condition, T(0)= 350

replacing = 280 into (1)

Hence, the differential equation becomes:

when;

time (t) = 1 hour

T(1) = 250

Since;

k = -0.4418

Therefore;

T(t) = 70 + 280e^{(-0.4418)}t

After 4 hours, the temperature is:

T(t) = 70 + 280e^{(-0.4418)}4

T(4) = 117.83° F