# The electric flux through a circle that lies in the xy-plane where electric field is E=(1200i...

## Question:

The electric flux through a circle that lies in the xy-plane where electric field is E=(1200i +500j +2500k)N/C is, 12.5 Nm{eq}^2 {/eq}/C. Find the diameter of the circle.

## Electric Flux:

In the region of electric field, the electric flux through any surface is the product of normal component of electric field with the area of the surface, which is also obtained by the dot product of electric field vector with the area vector of the surface.

Given Data

• Electric Field in the region is given as:
{eq}\vec E\ = 1200\hat i\ + 500\hat j\ + 2500\hat k\ N/C{/eq}
• A circle of diameter d (to be found) on x-y plane
• Electric Flux through the given circle, {eq}\phi\ = 12.5\ N.m^2/C{/eq}

Finding the electric flux ({eq}\phi{/eq}) through the circle

The area vector of the circle is expressed as:

• {eq}\vec A\ = \dfrac{\pi d^2}{4}\ \hat k{/eq}

The electric flux through the given circle is expressed as:

• {eq}\phi\ = \vec E\ \cdot \vec A{/eq}
• {eq}\phi\ = (1200\hat i\ + 500\hat j\ + 2500\hat k)\ \cdot (\dfrac{\pi d^2}{4}\ \hat k){/eq}
• {eq}12.5\ = 2500\times \dfrac{\pi d^2}{4}{/eq}
• {eq}d\ = 0.08\ m{/eq} 