Question

Define the terms electric field (E) and the potential difference between the two points. Derive the relationship between the two.

Answer 1

1. The electric field intensity at a point is the force experienced by a unit positive charge placed at that point.
2. The electric field intensity determines the strength of an electric field at that point.
3. The potential difference between two points informs us how much energy a unit charges obtains when traveling from one point to the next.
4. Let a charge q be moved from point A to point B against the electric field.
5. The force on q is $F=qE$
6. Then by the definition of potential difference, work done in moving the charge q from A to B is

$V_B-V_A=dV=\frac{W_{AB}}{q}$ …………(I)

7. Also $W_{AB}=F.dr=Fdr\cos \left(180\right)=-Fdr(wheredristhesmalldisplacementfromAtoB.$ …………(II)

8. From (i) and (ii). we get $$\begin{gathered} V_B-V_A=-\frac{Fdr}{q}=-\frac{qEdr}{q} \\ \Rightarrow dV=-EdrorE=-\frac{dv}{dr} \end{gathered}$$

Where,

V = Potential difference between the two points.

r = distance between the two points.