Question

The electric potential changes from $10\mathrm{V}$ to $30\mathrm{V}$ as one moves along $\mathrm{x}$-axis from $\mathrm{x}=0$ to $\mathrm{x}=2\mathrm{m}$. The magnitude of the electric field in the origin is

Answer 1

Step 1: Given data:

The electric potential changes from $10\mathrm{V}$ to $30\mathrm{V}$

The electric potential changes along $\mathrm{x}$-axis from $\mathrm{x}=0$ to $\mathrm{x}=2\mathrm{m}$

Step 2: Formula used:

Electric field intensity ($\mathrm{E}$) is equal to the negative rate of change of potential with respect to the distance ($\mathrm{dx}$) or it can be defined as the negative rate of derivative of potential difference ($\mathrm{dV}$).

$$\begin{gathered} \mathrm{dV}=-\overrightarrow{\mathrm{E}}\cdot \overrightarrow{\mathrm{dx}} \\ \mathrm{dV}=-\mathrm{Edxcos\theta } \\ \mathrm{E}=-\frac{\mathrm{dV}}{\mathrm{cos\theta dx}} \end{gathered}$$

Step 3: Electric field:

The magnitude of the electric field is,

$\mathrm{E}=-\frac{\mathrm{dV}}{\mathrm{cos\theta dx}}$

$\cos \mathrm{\theta }\approx 1$, if we take $\cos \mathrm{\theta }=1$,then

$\mathrm{E}=-\frac{\mathrm{dV}}{\mathrm{dx}}$

Where,

$\mathrm{dV}=30-10=20\mathrm{V}$

$\mathrm{dx}=2-0=2\mathrm{m}$

Therefore,

$\mathrm{E}=-\frac{20}{2}=-10\mathrm{V}/\mathrm{m}$

Therefore, the magnitude of the electric field in the origin is $10\mathrm{V}/\mathrm{m}$