Question

Define electromotive force of a cell.

Hint:

Remember the measurement rule: Open-circuit voltmeter reading at the terminals gives emf (\(I=0 $\Rightarrow$ V_{\text{terminal}}=\mathcal{E}\)). Under load, use \(V_{\text{terminal}}=\mathcal{E}-Ir\).

Answer 1

Core definition.
The electromotive force (emf) \(\mathcal{E}\) of a cell is the work done by the cell's non-electrostatic (chemical) forces in moving unit positive charge once around the entire circuit (including the cell's interior).
Mathematically, \[ \mathcal{E} = \frac{W}{Q} \] where \(W\) is the work performed by the cell on charge \(Q\). Its SI unit is the volt (V) \((= \text{J·C}^{-1})\).
Physical interpretation.
Inside the cell, chemical reactions separate charges against the internal electric field. This "source" work per unit charge is the emf. When no current flows (open circuit), the terminal potential difference equals \(\mathcal{E}\).
Circuit relation (with internal resistance).
If the cell has internal resistance \(r\) and supplies a current \(I\) to an external circuit of resistance \(R\), then the terminal voltage is \[ V_{\text{terminal}} = \mathcal{E} - I r $\Rightarrow$ \mathcal{E} = IR + Ir = I(R+r). \] Thus \(\mathcal{E}\) represents the "total available push" per unit charge; a portion \(Ir\) is lost inside the cell.
Integral form (general).
More generally, for any source, \[ \mathcal{E} = \oint \frac{\vec{f}_{\text{non-elec}}}{q}\cdot d\vec{\ell}, \] the line integral of the non-electrostatic force per unit charge around the loop (chemical, mechanical, photovoltaic, etc.).
Key properties.
\(\mathcal{E}\) is independent of load in the ideal (no internal resistance) case; real cells show a drop in terminal voltage under load due to \(r\).
Dimensions: \([\mathcal{E}] = \text{J·C}^{-1} = \text{kg·m}^2\text{·s}^{-3}\text{·A}^{-1}\) (volt).