Physics, 31.07.2019 13:30 hdhshshs741

Acapacitor with capacitance (c) = 4.50 μf is connected to a 12.0 v battery. what is the magnitude of the charge on each of the plates?

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Answer from: 573589

The charge on each plates is $54\mu C$ .

Explanation:

The magnitude of charge that flows through the plates is directly dependent on the capacitance and the voltage flowing through the plates.

Mathematically,

Q=CV

where,

Q = charge flowing = ? C

C = capacitance = $4.5\mu F=4.5\times 10^{-6}F$ (Conversion Factor: $1\mu F=10^{-6}F$ )

V = Voltage across the plates = 12 V

Putting values in above equation, we get:

$Q=4.5\times 10^{-6}F\times 12V=54\times 10^{-6}C=54\mu C$

Hence, the charge on each plates is $54\mu C$ .

Answer from: dhernandez081

The magnitude of the charge on each plates of capacitor is $\boxed{54 \times {10^{ - 6}}\,{\text{C}}}$ or $\boxed{54.0\mu \text{F}}$ .

Further Explanation:

When a capacitor is connected to a battery, the capacitor gets charged immediately. Charging takes no time. Positive charge is accumulated on the plate which is connected through a positive terminal of a battery and negative charge of the same magnitude is accumulated on the plate which is connected to the negative terminal of battery.

Given:

The capacitance of the capacitor is $4.50\mu\text{F}$ or $4.50 \times {10^{ - 6}}\,{\text{F}}$ .

The EMF of battery is $12.0\,{\text{V}}$ .

Concept:

Flow of charge in a circuit is called the electric current. Capacitor is used to store and supply the charge whenever it is required. When capacitor is connected through a battery terminal, electric current will flow into the capacitor and charge gets accumulated on the plates as the charge cannot pass through the insulated dielectric.

The charge accumulated on the capacitor is given by the following relation .

$\boxed{Q=CV}$ …… (1)

Here, is the charge accumulated on the capacitor plate, is the capacitance of the capacitor and is the voltage across the capacitor.

Substitute $4.50\mu\text{F}$ for and $12.0\text{ V}$ for in equation (1) .

$\begin{aligned}\\Q&=(4.5\times10^{-6}\text{C})(12.0\text{ V})\\&=54\times10^{-6}\text{C}\\\end{aligned}$

Thus, the magnitude of the charge on each plates of capacitor is $\boxed{54 \times {10^{ - 6}}\,{\text{C}}}$ or $\boxed{54.0\mu \text{F}}$ .

Learn more:

1. The motion of a body under friction

2. A ball falling under the acceleration due to gravity

3. Conservation of energy

Answer Details:

Grade: High school

Subject: Physics

Chapter: Capacitor

Keywords:

Capacitor, capacitance, 4.50 times 10^-6 F, 4.50 micro farad, 4.50 uF, 12.0 V, battery, magnitude, charge, plates, 54.0 times 10^-6 C, 54 times 10^-6 C, 54.0 micro coulomb, 54 micro coulombs, 54 uC, 54.0 uC.

Acapacitor with capacitance (c) = 4.50 μf is connected to a 12.0 v battery. what is the magnitude of

Answer from: Quest

Ithink the answer for this is: an electromagnet is a magnet with a soft metal core.

Answer from: Quest

just took the test its not "d"

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Acapacitor with capacitance (c) = 4.50 μf is connected to a 12.0 v battery. what is the magnitude of...