The amount of charge that is stored in a 3 μf capacitor, when the provided voltage is 220 V, is 6.6 m coulomb. This is obtained by the formula, Q=CV, where the Q is denoted as the amount of charge, C equals to the value of capacitance, and V is regarded as the voltage difference between the two spaced plates of the capacitor.

This capacitor works by building up opposite charges on parallel plates when a voltage is applied from one plate to the other. The amount of charge that moves into the plates depends upon the capacitance and the applied voltage according to the formula Q=CV.

# How to Measure charge on 3.0 μf capacitor?

A capacitor is used to store electric charge. The more voltage (electrical pressure) we apply to the capacitor, the more charge is accumulated within the capacitor. Also, the more capacitance the capacitor possesses, the more charge will be forced in by a given voltage.

As capacitance defines the capacitors ability (capacity) to store an electrical charge on its plates. We can state one Farad as the “**capacitance of a capacitor which requires a charge of one coulomb to establish a potential difference of one volt between its plates**” as firstly described by Michael Faraday. So the more the capacitance, the higher is the amount of charge stored on a capacitor for the same amount of voltage.

### What is the Charge of the Capacitor 2.0 μf?

A 2 μF capacitors can withstand a potential difference of not more than 300V.

The formula for obtaining the charge of the capacitor is **Q=CV**

Hence, the charge obtained by the capacitor is, **Q= (2*10 ^{-6}*300) = 0.6 m Coulomb**.

### What is the Potential Difference Across the Plates of the 6.0-μf Capacitor?

For an amount of 6-μf capacitor, if the capacitor keeps an amount of 10 Joules of energy within itself, then the voltage across the capacitor will be,

**V _{1}=√{10/(6*10^{-6})}= 1291 Volts**.

Again, for an amount of 6-μf capacitor, if the capacitor keeps an amount of 0.1 Joules of energy within itself, then the voltage across the capacitor will be, **V _{1}=√{0.1/(6*10^{-6})}= 129 Volts**.

## What is the energy stored by the capacitor 2.0-μf?

The formula for obtaining the amount of energy stored by the capacitor is **E=CV ^{2}**

The amount of energy stored in a 2-μf capacitor when the provided voltage difference between the capacitor plates is 220 Volts, then the amount of energy will be

**E= (2*10 ^{-6}*220^{2}) Volts= 0.2904 Volts**.

## Conclusion

Low voltage super-capacitors are commonly used in portable hand held devices to replace large, expensive and heavy lithium type batteries as they give battery-like storage and discharge characteristics making them ideal for use as an alternative power source or for memory backup. Super-capacitors used in hand held devices are usually charged using solar cells fitted to the device.

Ultra-capacitor are being developed for use in hybrid electric cars and alternative energy applications to replace large conventional batteries as well as DC smoothing applications in vehicle audio and video systems. Ultra-capacitors can be recharged quickly and have very high energy storage densities making them ideal for use in electric vehicle applications.

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