kyle696969
20.07.2020 •
Physics
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
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Ответ:
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
Ответ:
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.
Ответ:
Explanation:
(4566 m / 4 min) × (1 km / 1000 m) × (60 min / h) = 68.49 km/h