Question

A 1.97-pF capacitor is connected to a 9.0-V battery and fully charged. How many electrons did the battery transfer from one capacitor plate to the other

Answer 1

Answer:

1.11×10⁸  Electrons.

Explanation:

Applying,

Q = CV..................... Equation 1

Where Q = Total charge on the capacitor, C = Capacitance of the capacitor, V = Voltage of the battery.

Given: C = 1.97 pF = 1.97×10⁻¹² F, V = 9.0 V

Substitute this value into equation 1

Q = 1.97×9×10⁻¹²

Q = 1.773×10⁻¹¹ C.

If the charge on one electron = 1.602×10⁻¹⁹ C.

Therefore number of electron = 1.773×10⁻¹¹ /1.602×10⁻¹⁹

Number of electron = 1.11×10⁸

Answer 2

Answer:

The battery transferred 1.107 x 10⁸ electrons from one capacitor plate to the other.

Explanation:

Given;

capacitance of the capacitor, C = 1.97 pF = 1.97 x 10⁻¹² F

the battery potential, V = 9 V

The charge on each capacitor is calculated as;

Q = CV

Q = 1.97 x 10⁻¹² x 9

Q = 1.773 x 10⁻¹¹ C

1 electron has 1.602 x 10⁻¹⁹ Coulomb's charge

⇒1.602 x 10⁻¹⁹ C = 1 electron

⇒1.773 x 10⁻¹¹ C = ?

= (1.773 x 10⁻¹¹ C x electron) / (1.602 x 10⁻¹⁹ C)

= 1.107 x 10⁸ electrons

Therefore, the battery transferred 1.107 x 10⁸ electrons from one capacitor plate to the other.