Question

A 0.160 pF parallel-plate capacitor is charged to a potential difference of 10.0 V and then disconnected from the battery. A cosmic ray burst creates 1.00 x 106 electrons and 1.00 x 106 positive charges between the plates. If the charges do not recombine, but reach the oppositely charged plates, by how much is the potential difference between the capacitor plates reduced

Answer 1

The potential difference across a 0.160 pF parallel-plate capacitor is reduced by 1.00 V when it is charged with 1.00 x 10^6 electrons, resulting in a new potential difference of 9.00 V.

Step-by-step explanation:

The question involves a scenario where a 0.160 pF parallel-plate capacitor is initially charged to a 10.0 V potential difference and then isolated. A cosmic ray burst then creates equal amounts of electrons and positive charges that are assumed to reach the opposite plates of the capacitor, which would affect the potential difference across the plates.

To calculate how much the potential difference is reduced, we need to determine the charge introduced by the cosmic rays and then relate this to the potential difference using the formula V = Q/C, where V is the potential difference, Q is the charge, and C is the capacitance of the capacitor.

Each electron has a charge of approximately 1.60 x 10^-19 C. Therefore, the total charge introduced by the electrons and positive charges is 1.00 x 10^6 electrons x 1.60 x 10^-19 C/electron = 1.60 x 10^-13 C. The potential difference reduced by this additional charge is ΔV = Q/C = (1.60 x 10^-13 C) / (0.160 x 10^-12 F) = 1.00 V.

Therefore, the new potential difference across the capacitor plates after the cosmic ray burst would be 10.0 V - 1.00 V = 9.00 V.

Answer 2

1.0 volts

Step-by-step explanation:

Given that:

The potential difference between the plate V = 10 V

C = 0.160 pF = 0.160 × 10⁻¹² F

The charge on the capacitor

Q = CV

Q = 0.160 × 10⁻¹² × 10

Q = 1.6 × 10⁻¹² C

It implies that the positive plate of capacitor has +1.6 × 10⁻¹² C charge while the negative plate has -1.6 × 10⁻¹² C

The number of excess electrons on the negative plate is:

`n = (q)/(e)`

`n = (-1.6* 10^(-12))/(-1.6* 10^(-19))`

`n = 1.0 * 10^7`

Thus, electron deficiency on the positive plate is
`1.0 * 10^7`

The net negative charge that moves towards the positive plate is :

q = number of electrons moved × e

`q = 1* 10^6 * (-1.6 * 10^(-19))`

`q = -1.6 * 10^(-13) \ C`

Now, the net charge on the positive plate is:

`q_(net) = q +q' \\ \\ q_(net) = (1.6 * 10^(-12)) + (-1.6 * 10^(-13))`

`q_(net) =1.44 * 10^(-12) \ C`

The potential difference between the plate;

`V_(new) = (q_(net) )/(c)`

`V_(new) = (1.44 * 10^(-12) )/(0.16 * 10^(-12))`

`V_(new) = 9.0 V`

The reduction in potential difference

`\Delta V = V - V_(new)`

`\Delta V = 10 - 9.0`

`\mathbf{\Delta V = 1.0 \ volts}`