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Dimensional Formula of Electrical Capacitance

Capacitance is the ability of a system (like a capacitor) to store an electric charge. Finding its dimensional formula requires combining several fundamental concepts like Work, Charge, and Potential Difference.

Question (Click to Flip)

What factors determine the capacitance of a parallel plate capacitor?

Answer

It depends purely on the physical geometry: the surface area of the plates, the distance between them, and the dielectric material placed inside them.

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Key Facts

The SI unit of capacitance is the Farad (F), named after the English scientist Michael Faraday.

Because 1 Farad is an incredibly massive amount of capacitance, practical capacitors are usually measured in microfarads ($\mu F$) or picofarads (pF).

1. The Base Formula

The basic formula for Capacitance ($C$) is: Capacitance ($C$) = $\frac{\text{Charge (Q)}}{\text{Voltage (V)}}$

2. Finding Voltage (Potential Difference)

Voltage ($V$) is defined as Work Done ($W$) per unit Charge ($Q$).

  • Formula: $V = \frac{W}{Q}$
  • Dimension of Work = Force $\times$ Displacement = $[M^1 L^2 T^{-2}]$
  • Dimension of Charge = Current $\times$ Time = $[A^1 T^1]$
  • So, Dimension of Voltage ($V$) = $\frac{[M^1 L^2 T^{-2}]}{[A^1 T^1]} = [M^1 L^2 T^{-3} A^{-1}]$

3. The Final Derivation

Now, substitute the dimensions of Charge and Voltage back into the main Capacitance formula ($C = Q/V$):

$C = \frac{[A^1 T^1]}{[M^1 L^2 T^{-3} A^{-1}]}$

Bring all the terms from the denominator to the numerator (change the signs): $C = [A^1 T^1] \times [M^{-1} L^{-2} T^3 A^1]$

Combine the similar terms ($A^{1+1}$ and $T^{1+3}$): Final Dimensional Formula = $[M^{-1} L^{-2} T^4 A^2]$

Questions and Answers

What factors determine the capacitance of a parallel plate capacitor?+

It depends purely on the physical geometry: the surface area of the plates, the distance between them, and the dielectric material placed inside them.

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