Question

A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50 × 10−3 m. The plates of the capacitor are separated by a space of 1.40 × 10−4 m. If the potential difference of 0.12 V is removed from the circuit and the circuit is allowed to discharge until the charge on the plates has decreased to 70.7 percent of its fully charged value, what will the potential difference across the capacitor be in V?

Answer 1

The potential difference is
`V = 8.5 *10^(-2) \ volt`

Step-by-step explanation:

From the question we are told that

The diameter of the of both plates is
`D = 2.50 *10^(-3) \ m`

The distance of separation is
`d = 1.40 *10^(-4) \ m`

The potential difference is
`V = 0.12 \ V`

The new charge on the plate is
`q = 70.7`% of
`Q_o`

Where
`Q_o` is the original charge on the capacitor

Generally the charge on the capacitor is mathematically represented as

`Q = CV`

Where C is the capacitance of the capacitor which is mathematically represented as

`C = \epsilon_o (A)/(d)`

Where
`\epsilon_o` is the permittivity of free space which is a constant with value

`\epsilon_o = 8.85 *10^(-12) F/m`

A is the area which is mathematically represented as

`A = \pi (D^2)/(4)`

substituting values

`A = 3.142 * (6.25*10^(-6))/(4)`

`A = 4.909*10^(-6) \ m^2`

Now

`C = 8.85 *10^(-12) (4.909 *10^(-6))/(1.40 *10^(-4))`

`C = 3.1014 *10^(-13) \ F`

So

`Q_o = 3.1014 *10^(-13)* 0.12`

`Q_o = 3.72 *10^(-14) \ C`

Generally

`q= CV`

Here

`q = (70.7)/(100) * Q_o`

`q = (70.7)/(100) * 3.72*10^(-14)`

`q = 2.636*10^(-14) \ C`

Making V the subject in the above formula

`V = (q)/(C)`

Substituting values

`V = (2.636 *10^(-14))/(3.1014 *10^(-13) )`

`V = 8.5 *10^(-2) \ volt`