Question

Estimate the electric fields ${\overrightarrow{E}}_1\mathrm{and}{\overrightarrow{E}}_2$at points 1 and 2 in Figure Q26.5. Don’t forget that $\overrightarrow{E}$is a vector.

Short answer

The electric field varies in the direction 1 to 2.

$$\begin{gathered} {\overrightarrow{E}}_1=-5\mathrm{V}/\mathrm{m} \\ {\overrightarrow{E}}_2=-20\mathrm{V}/\mathrm{m} \end{gathered}$$

Step-by-step solution

Given information and formula used   

Given :

Theory used :

Electric Field is given by :$E=\frac{V}{r}$

Estimating E→1 and E→2

(a) We can see that the potential difference in $2metres\mathrm{is}10Volts$, hence the field strength at this point will be $5V/m$. The formula is as follows:

localid="1650261683495" $\begin{array}{rcl}{\overrightarrow{E}}_1& =& \frac{-10+0}{2}\\ & =& \boxed{\mathbf{-}\mathbf{5}\mathbf{}\mathit{V}\mathbf{/}\mathit{m}}\end{array}$

(b) We can see that the potential difference in $1metre\mathrm{is}20Volts$, hence the field strength at this location will be $20V/m$. The formula is as follows:

localid="1650261669941" $\begin{array}{rcl}{\overrightarrow{E}}_2& =& \frac{-(40-20)}{1}\\ & =& \boxed{\mathbf{-}\mathbf{20}\mathbf{}\mathit{V}\mathbf{/}\mathit{m}}\end{array}$

The vector will point to the left in both circumstances, parallel to the $x$-axis.