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202209281050 Exercise 22.2 (Q23)

An electric field is given by \overrightarrow{E}=E_0\,\hat{\jmath}, where E_0 is a constant. Find the potential as a function of position, taking V=0 at y=0.

Extracted from R. Wolfson. (2016). Essential University Physics.

Background.

Electric field vector \overrightarrow{E} is related to electric potential V by:

\displaystyle{\overrightarrow{E}= -\bigg( \frac{\partial V}{\partial x}\,\hat{\mathbf{i}} + \frac{\partial V}{\partial y}\,\hat{\mathbf{j}} + \frac{\partial V}{\partial z}\,\hat{\mathbf{k}} \bigg)}

or reversely,

\displaystyle{V(r)=-\int\overrightarrow{E}\cdot\mathrm{d}\vec{r}}.

Write

\displaystyle{(0,E_0,0) = \bigg( -\frac{\partial V}{\partial x},-\frac{\partial V}{\partial y},-\frac{\partial V}{\partial z}\bigg) }

that implies V=V(y) is x– and z-independent.

Hence

\begin{aligned} V(y) & = -\int E_0\,\mathrm{d}y \\ & = -E_0y+C\qquad \textrm{for some constant }C \\ \because\enspace 0 & = V(y=0) = -E_0(0)+C \\ \Rightarrow C& = 0 \\ \therefore\enspace V&=V(y)=-E_0y\\ \end{aligned}