Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

This question is for aspirants: The electric field strength depends only on the x and y coordinates according to the law where a is a constant, i and j are unit vectors along x and y axes respectively. Find the flux of the vector E through a sphere of radius R with its centre at the origin of coordinates. its a simple problem.

This question is for aspirants:

The electric field strength depends only on the x and y coordinates according to the law

   



            



 



where a is a constant, i and j are unit vectors along x and y axes respectively. Find the flux of the vector E through a sphere of radius R with its centre at the origin of coordinates.

its a simple problem.

Grade:Upto college level

2 Answers

aman walia
24 Points
11 years ago

 respected sir

 

 i m little confused about this question. actually electric field is 2-d only but for anwsering flux we have to consider total surface area.

 

well i am presenting my solution:;

 

E at (R,0,0) is a/R. similarly E at (0,R,0) is also a/R. so this could be taken as the a simple charge system such that a charge q is placed at origin.

 

now

kq/R^2 =a/R

thus q=1/k* a*R

so flux is 4pi*a*R.

i am not sure this is right . if u find it wrong plz correct it.

Akash Verma
36 Points
11 years ago

pleeeez tell d answer b'coz the answer coming is radius or size dependent which is not possible

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free