Question

Show that \(\frac{\text{Farad}}{\text{meter}} = \frac{\text{coulomb}^2}{\text{newton} \times \text{meter}^2}\). Name its physical quantity. If potential difference across ends of capacitor of capacitance 6 \(\mu\)F is 2 volts, find out the potential difference across ends of the battery :
\begin{center} \begin{circuitikz} \draw (0,2) to[C, l=4F] (2,2) to[C, l=6F] (4,2); \draw (4,2) -- (4,0); \draw (4,0) to[battery1, l=V] (0,0); \draw (0,0) -- (0,2); \end{circuitikz} \end{center}

Hint:

For capacitors in series, the charge is the same on all of them. This is the key piece of information. Calculate the charge on the one capacitor where you have enough information (C and V), and then use that charge value for all other capacitors in the series branch.

Answer 1

Part 1: Unit Derivation

Step 1: Understanding the Concept:
We need to show the equivalence of two sets of units by breaking them down into their fundamental definitions.

Step 2: Derivation:
Let's start with the Left Hand Side (LHS): \(\frac{\text{Farad}}{\text{meter}}\).
By definition, capacitance \(C = Q/V\), so the unit Farad (F) is equivalent to Coulomb per Volt (C/V). \[ \text{LHS} = \frac{\text{C/V}}{\text{m}} = \frac{\text{C}}{\text{V} \cdot \text{m}} \] The unit for potential difference, Volt (V), is defined as energy per unit charge, which is Joule per Coulomb (J/C). \[ V = \frac{J}{C} \] The unit for energy, Joule (J), is defined as work done, which is force times distance, or Newton-meter (\(N \cdot m\)). \[ J = N \cdot m \] Substituting these back into our expression for the LHS: \[ V = \frac{N \cdot m}{C} \] Now, substitute this expression for Volt into the LHS: \[ \text{LHS} = \frac{\text{C}}{\left(\frac{N \cdot m}{C}\right) \cdot \text{m}} = \frac{\text{C}^2}{N \cdot m^2} \] In terms of base units, this is \(\frac{\text{coulomb}^2}{\text{newton} \times \text{meter}^2}\), which is exactly the Right Hand Side (RHS). Thus, the relation is proven.
Part 2: Name the Physical Quantity
The unit \(\frac{\text{Farad}}{\text{meter}}\) is the SI unit for **electric permittivity (\(\epsilon\))**. The permittivity of free space, \(\epsilon_0\), has a value of approximately \(8.854 \times 10^{-12}\) F/m.
Part 3: Circuit Problem

Step 1: Understanding the Circuit:
The circuit shows two capacitors, \(C_1 = 4 \, \mu\text{F}\) and \(C_2 = 6 \, \mu\text{F}\), connected in series to a battery of voltage \(V\).

Step 2: Key Principles for Series Capacitors:
\begin{enumerate} \item The charge stored on each capacitor is the same: \(Q_1 = Q_2 = Q_{total}\). \item The total voltage of the battery is the sum of the voltages across each capacitor: \(V = V_1 + V_2\). \end{enumerate}

Step 3: Calculation:
We are given the potential difference across the \(6 \, \mu\text{F}\) capacitor: \[ V_2 = 2 \, \text{V} \] First, we find the charge stored on this capacitor using \(Q = CV\): \[ Q_2 = C_2 V_2 = (6 \, \mu\text{F})(2 \, \text{V}) = 12 \, \mu\text{C} \] Since the capacitors are in series, the charge on the \(4 \, \mu\text{F}\) capacitor is the same: \[ Q_1 = Q_2 = 12 \, \mu\text{C} \] Now, we can find the voltage across the \(4 \, \mu\text{F}\) capacitor: \[ V_1 = \frac{Q_1}{C_1} = \frac{12 \, \mu\text{C}}{4 \, \mu\text{F}} = 3 \, \text{V} \] The total potential difference across the battery is the sum of the individual voltages: \[ V = V_1 + V_2 = 3 \, \text{V} + 2 \, \text{V} = 5 \, \text{V} \]

Step 4: Final Answer:
The potential difference across the ends of the battery is 5 V.