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Travka [436]
4 years ago
7

Why is there no electric field at the center of a charged spherical conductor?

Physics
1 answer:
Svetlanka [38]4 years ago
4 0
-- Since the sphere is a conductor, the charge on it will move around
until it's evenly distributed on the surface of the sphere.  When every
tiny smitch of charge is the same distance from the charge around it,
they won't need to move around any more.

-- At that point, the situation at the center of the sphere will be:
For every smitch of charge on the surface, causing en electric field
at the center, there is another smitch, of the same size and in exactly
the opposite direction, canceling out the field of the first one. 
Every smitch of charge on the surface causes a tiny bit of electric field
at the center, and they all cancel each other.

It turn out that if the sphere is hollow, then the electric field at EVERY
point inside it is zero, not only at the center.

It's exactly the same idea as a sphere with uniform, homogeneous mass.
For that sphere, the gravity at the center is zero.
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A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.
kodGreya [7K]

Answer:

The  z-component of the force is  \= F_z  =  0.00141 \ N    

Explanation:

From the question we are told that

          The charge on the particle is q =  7.76 *0^{-8} \  C    

           The magnitude of the magnetic field is  B =  0.700\r i \ T

            The  velocity of the particle toward the x-direction is  v_x  =  -1.68*10^{4}\r  i  \ m/s

           The  velocity of the particle toward the y-direction is

v_y  =  -2.61*10^{4}\ \r j  \ m/s

           The  velocity of the particle toward the z-direction is

v_y  =  -5.85*10^{4}\ \r k  \ m/s

Generally the force on this particle is mathematically represented as

          \= F  =  q (\= v   X  \= B )

So  we have    

          \= F  =  q ( v_x \r  i + v_y \r  j  +  v_z \r k  )  \ \ X \ (  \= B i)

         \= F  = q (v_y B(-\r  k) + v_z B\r j)      

  substituting values

       \= F  = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r  z) + [(5.58*10^{4}) (0.700)]\r y)    

      \= F=  0.00303\ \r j +0.00141\ \r k                  

So the z-component of the force is  \= F_z  =  0.00141 \ N    

Note :  The  cross-multiplication template of unit vectors is  shown on the first uploaded image  ( From Wikibooks ).

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