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3 years ago

11

Consider the uniform electric field \vec{E} =(4.00~\hat{j}+3.00~\hat{k})\times 10^3~\text{N/C} E ⃗ =(4.00 j ^ +3.00 k

^ )×10 3 N/C. What is its electric flux through a circular area of radius 2.66 m that lies in the xy-plane?

1 answer:

Tema [17]3 years ago

6 0

Answer:

Electric flux, $\phi=6.668\times 10^4\ Nm^2/C$

Explanation:

It is given that,

Electric field, $E=(4j+3k)\times 10^3\ N/C$

We need to find the electric flux through a circular area of radius 2.66 m that lies in the xy-plane. $A=Ak$

The electric flux is given by :

$\phi=E{\cdot}A$

$\phi=(4j+3k)\times 10^3{\cdot}Ak$

Since, k.k=i.i=j.j = 1

So,

$\phi=3\times 10^3\times \pi\times (2.66)^2\ k$

$\phi=6.668\times 10^4\ Nm^2/C$

So, the electric flux through a circular area is $\phi=6.668\times 10^4\ Nm^2/C$ .Hence, this is the required solution.

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