Question

The variation of electric potential $\left(V\right)$ with distance $\left(r\right)$ from a fixed point is as shown in the figure. Electric field intensity $(\text{in V/m)}$ at $r=1\text{m},3\text{m}$ and $4.5\text{m}$ are respectively

• A. $+8,-1,-\frac{4}{3}$
• B. $2,0,-\frac{8}{3}$
• C. $-2,1,\frac{8}{3}$
• D. $-2,0,\frac{8}{3}$

Answer 1

The correct option is D $-2,0,\frac{8}{3}$

Given figure:


The relation between electric field and potential is given by
$E=-\frac{\partial V}{\partial r}$

That means slope of $V-r$ graph gives electric field $E$.

So,
Case $\left(i\right):$ At $r=1\text{m}$
Electric field $E_1=$Slope $\text{OA}=-\frac{\Delta V}{\Delta r}$

Substituting the values, we get

$E_1=-\frac{(4-0)}{(2-0)}$

$\therefore E_1=-2{\text{Vm}}^{-1}$

Case $\left(ii\right):$ At $r=3\text{m}$

Slope $\text{AB}=0$

$\therefore E_2=0\text{V/m}$

Case $\left(iii\right):$ At $r=4.5\text{m}$

Slope of $\text{BC}=E_3=-\left[\frac{0-4}{5.5-4}\right]$

$\therefore E_3=\frac{4}{1.5}=\frac{8}{3}\text{V/m}$

Hence, option $\left(\text{B}\right)$ is the correct answer.