Not always a sphere!

Electricity and Magnetism Level 5

The electric field strength depends only on the x x and y y coordinates according to the law E ⃗ = a ( x i ^ + y j ^ ) / ( x 2 + y 2 ) \vec{E} = a(x\hat{i} + y\hat{j})/(x^2 + y^2) , where a a is a constant.

Find the flux of the vector E ⃗ \vec{E} through the closed surface x 4 + y 4 + z 4 = 81 x^4 + y^4 + z^4 = 81 .

If it can be expressed k π a k\pi a then give the value of ⌊ 100 k ⌋ \left \lfloor{100k} \right \rfloor .

Notation : ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is 1200.00.

Rajdeep Dhingra by Rajdeep Dhingra , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

The electric field given is that of an infinite line charge of uniform density along the z axis. The given surface encloses the portion |z| < 3 of the line charge. Using Gauss' Law, we can relate the total charge enclosed, to the electric flux. This will yield an answer of 1200.

15 Helpful 1 Interesting 1 Brilliant 0 Confused

You got it. Great Job ! First approach of many people is to integrate.

Rajdeep Dhingra - 4 years, 10 months ago

I tried solving this by integration. Taking out the surface normal by gradient of the surface then doing surface integration. The integration was so complicated! How did you come up with this approach? Can you give me any reference to such kind of problems?

Nashita Rahman - 2 years, 5 months ago
Alapan Chaudhuri
May 23, 2018

The Electric field expression is independent of z z . Also it depends on x x and y y symmetrically. From this we may infer that the given charge distribution must be along z axis and such that it radially gives out fields. We can reexpress Electric field expression as a / r a/r in the r-cap direction where r is the linear distance of the point from z axis. This is equivalent to the electric field intensity expression of a infinite line of charge with uniform charge density. Thus by the uniqueness principles the given distribution is an infinite line charge along z axis.

Now since the given surface has already been defined we know the extent of the infinite line within the surface which is 6 units, from +3 to -3 along z axis (in units). This yields the required flux q / ε q/ε where q = λ l q=λl and thus q / ε = λ l / ε = 4 π λ l / 4 π ε = 2 l π a q/ε=λl/ε=4πλl/4πε=2lπa .

Thus, k = 2 l = 12 k=2l=12 and hence the required answer is 1200.

1 Helpful 1 Interesting 1 Brilliant 1 Confused

How can you express electric field as a/r ? Can you explain me how you're using uniqueness principles here?

Nashita Rahman - 2 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...