MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
umesh chandrakant shah Grade: 12
        The electric field strengh in region is given as E=Xi+Yj / X^2+Y^2. Find the net charge inside a sphere of radius 'a' with its center at origin. (a=2M)write value of 'k' if charge is k epsulon.
7 years ago

Answers : (1)

Pratham Ashish
17 Points
										

consider the sphere as a gaussian surface then,


    § E . ds  = Q in /£ ..........................(1)


   =  Xi+Yj / X^2+Y^2   at any point (x,y) on the surface of the sphere ,


 area vector is always parallel to the surface, so in the case of the sphere it


will be in the direction of radius vector ,


so,  at any point (x,y)  the area vector  Xi+Yj  /  (X^2+Y^2) 1/2  ds ,


                 where  Xi+Yj  /  (X^2+Y^2)  is the unit vector in the direction of area vector 


so the l.h.s of eq (1),


      § E . ds   =


                       = §(Xi+Yj / X^2+Y^2) .{ Xi+Yj  /  (X^2+Y^2) 1/2 }  ds


                    =  § (X^2+Y^2) / (X^2+Y^2) 3/2 ds


                    = §1/ (X^2+Y^2) 1/2  ds


 


( for a sphere  (X^2+Y^2) 1/2  =  radius = 2M)


                   = 1/2M  §ds


         (   total surfase area for a sphere =  4 pi r^2


                                                                  = 4 pi * 4 M^2)


                        1/2M  * 16 *pi * M^2


                     =  8 pi m^3 =   r.hs.


so


  8 pi m^3  =  Q in


 Q = 8 pi m^3 £


           k= 8 pi m^3


 


 

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: R 2,600
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: R 2,800
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details