What is the Investigation about!
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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
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