If a capacitor stores 0.12 C at 10 V, then its capacitance is-
0.012 F
Capacitance is a fundamental electrical property that measures a component's ability to store an electric charge. A capacitor is a device specifically designed for this purpose. The relationship between the charge (\(Q\)) stored on a capacitor, the voltage (\(V\)) across it, and its capacitance (\(C\)) is given by a simple formula.
The question provides us with the following information:
We need to find the capacitance (\(C\)).
The relationship between charge, voltage, and capacitance is:
\begin{equation*} Q = C \times V \end{equation*}
To find the capacitance \(C\), we can rearrange this formula:
\begin{equation*} C = \frac{Q}{V} \end{equation*}
Now, we substitute the given values of \(Q\) and \(V\) into this formula:
\begin{equation*} C = \frac{0.12 \text{ C}}{10 \text{ V}} \end{equation*}
Performing the division, we get:
\begin{equation*} C = 0.012 \text{ Farads} \end{equation*}
The unit of capacitance is the Farad (F).
Therefore, the capacitance of the capacitor is 0.012 F.
| Concept | Definition/Formula | Unit |
|---|---|---|
| Charge (\(Q\)) | Fundamental property of matter that experiences a force when in the presence of an electromagnetic field. | Coulomb (C) |
| Voltage (\(V\)) | Electric potential difference between two points. Energy per unit charge. | Volt (V) |
| Capacitance (\(C\)) | Ability of a body to store an electrical charge. Ratio of the amount of electric charge stored on a conductor to a difference in electric potential. | Farad (F) |
| Formula relating Q, C, V | \(Q = C \times V\) | N/A |
Capacitors are essential components in electronic circuits. They can store electrical energy in an electric field between two conductive plates separated by an insulating material called a dielectric. Here are a few related points:
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