Question

In Figure 21.15 the magnitude of the electric field is $\mathbf{1000}\mathbf{ }\mathbf{\text{V/m}}$, and the field is at an angle of ${\mathbf{30}}^{\mathbf{o}}$to the outward-going normal. What is the flux on the small rectangle whose dimensions are $\mathbf{1}\mathbf{ }\mathbf{\text{mm}}$by$\mathbf{\text{2}}\mathbf{ }\mathbf{\text{mm}}$ ?

Short answer

$1.732\times {10}^{-3} \text{Vm}$

Step-by-step solution

Identification of given data

Magnitude of the electric field is $\left|\overrightarrow{E}\right|=1000 \text{V/m}$

Area of rectangle is $\Delta A=1 \text{mm}\times 2 \text{mm} \text{or} \text{2} \times {10}^{-6}{\text{m}}^2$

The field is at an angle of$\theta ={30}^o$

Significance of electric flux

The number of electric lines of force (or electric field lines), a characteristic of an electric field, that cross a specific area is known as electric flux.

$\varphi =\left|\overrightarrow{E}\right|\times \Delta A\times \cos \theta$

Determining the flux on the small rectangle whose dimensions are1 mm by 2 mm

Using equation (i)

$\begin{array}{rcl}\varphi & =& \left|\overrightarrow{E}\right|\times \Delta A\times \cos \theta \\ & =& 1000 \text{V/m}\times \text{2} \times {10}^{-6}{\text{m}}^2\times \cos {30}^o\\ & =& 1.732\times {10}^{-3} \text{Vm}\end{array}$

Hence the flux on the small rectangle is$1.732\times {10}^{-3} \text{Vm}$