Question

A capacitor 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?

Answer 1

Answer: 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 2

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, $Q$ is the charge accumulated on the capacitor plate, $C$ is the capacitance of the capacitor and $V$ is the voltage across the capacitor.  

Substitute $4.50\mu\text{F}$ for $C$ and $12.0\text{ V}$ for $V$ 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 brainly.com/question/4033012  

2. A ball falling under the acceleration due to gravity brainly.com/question/10934170

3. Conservation of energy brainly.com/question/3943029

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.