Apower source of 2.0 v is attached to the ends of a capacitor. the capacitance is 4.0 μf. what is the amount of charge stored in this capacitor? 0.50 μc 2.0 μc 6.0 μc 8.0 μc

Q = 8 μC

Explanation:
The relation between voltage, capacitance and charge can be expressed using the following rule:
Q = C * V
where:
Q is the amount of charge that we want to calculate
C is the capacitance = 4 * 10⁻⁶ F
V is the voltage applied = 2 V

Substitute with the givens in the above equation to get the amount of charge as follows:
Q = C * V
Q= 4 * 10⁻⁶ * 2
Q = 8 * 10⁻⁶ Coulumb
Q = 8 μC

Hope this helps :)

Voltage across capacitor, V = 2.0V
Capacitance of Capacitor, C = 4.0 μF
Charge stored in Capacitor, Q = ?

The capacitance formula, C = Q/V
We can rewrite this as Q = CV

Therefore, the amount of charge stored in this capacitor = 2.0 * 4.0 μC
= 8.0 μC

An atom that has fewer electrons that protons is called a positive ion.