Calculating Capacitor Capacitance from Charge and Voltage

The relationship between the charge (\(Q\)) stored on a capacitor, the voltage (\(V\)) across it, and its capacitance (\(C\)) is a fundamental concept in electrostatics. This relationship is given by the formula:

\[Q = CV\]

This formula tells us that the charge stored on a capacitor is directly proportional to both its capacitance and the voltage applied across it.

In this question, we are given the following information:

• The voltage supply (\(V\)) = 50 V
• The charge stored on the capacitor (\(Q\)) = 10 μC

We need to find the capacitance (\(C\)) of the capacitor. To do this, we can rearrange the formula \(Q = CV\) to solve for \(C\):

\[C = \frac{Q}{V}\]

Before substituting the values into the formula, it is important to ensure that the units are in the standard SI system. The voltage is already in volts (V), which is the SI unit for electric potential difference. The charge is given in microcoulombs (μC). We need to convert microcoulombs to Coulombs (C), which is the SI unit for electric charge. The conversion is:

\[1 \mu C = 10^{-6} C\]

So, the charge of 10 μC is equivalent to:

\[Q = 10 \times 10^{-6} C\]

Now we can substitute the values of \(Q\) and \(V\) into the rearranged formula for capacitance:

\[C = \frac{10 \times 10^{-6} C}{50 V}\]

Performing the calculation:

\[C = \frac{10}{50} \times 10^{-6} \frac{C}{V}\]

\[C = 0.2 \times 10^{-6} \frac{C}{V}\]

The unit C/V is equivalent to Farads (F), which is the SI unit for capacitance. So, the capacitance is:

\[C = 0.2 \times 10^{-6} F\]

The options are given in microfarads (μF). Since \(10^{-6} F = 1 \mu F\), we can convert our result to microfarads:

\[C = 0.2 \mu F\]

Thus, the capacitance of the capacitor is 0.2 μF.

Summary of Calculation Steps

1. Identify the known quantities (Charge Q, Voltage V) and the unknown quantity (Capacitance C).
2. Recall the formula relating these quantities: \(Q = CV\).
3. Rearrange the formula to solve for capacitance: \(C = \frac{Q}{V}\).
4. Ensure all given values are in standard SI units. Convert 10 μC to 10 x 10^-6 C.
5. Substitute the SI values into the formula and calculate C.
6. Convert the result back to the required unit (μF) if necessary.

Revision Table: Key Concepts in Capacitance Calculation

Additional Information on Capacitors and Capacitance

A capacitor is an electronic component designed to store electrical energy in an electric field. It typically consists of two conductive plates separated by a dielectric material (an insulator).

• How Capacitors Work: When a voltage is applied across the capacitor, charge builds up on the plates. One plate accumulates positive charge and the other negative charge, with the amount of positive charge equal to the amount of negative charge. The dielectric material increases the amount of charge that can be stored for a given voltage.
• Units of Capacitance: The SI unit of capacitance is the Farad (F). However, a Farad is a very large unit. Most capacitors used in electronic circuits have capacitances in the range of microfarads (μF), nanofarads (nF), or picofarads (pF).
  • 1 μF = 10^-6 F
  • 1 nF = 10^-9 F
  • 1 pF = 10^-12 F
• Energy Stored in a Capacitor: The energy (E) stored in a charged capacitor is given by several equivalent formulas: \[E = \frac{1}{2} CV^2 = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} QV\] The energy is stored in the electric field within the dielectric material.
• Factors Affecting Capacitance: The capacitance of a parallel-plate capacitor depends on the area (\(A\)) of the plates, the distance (\(d\)) between the plates, and the permittivity (\(\epsilon\)) of the dielectric material between the plates. The formula for a parallel-plate capacitor is: \[C = \frac{\epsilon A}{d}\] where \(\epsilon = \epsilon_r \epsilon_0\), \(\epsilon_r\) is the relative permittivity of the dielectric, and \(\epsilon_0\) is the permittivity of free space.