The electric field everywhere on the surface of a thin, spherical shell of radius 0.735 m is of magnitude 868 N/C and points radially toward the center of the sphere.

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The electric field everywhere on the surface of a thin, spherical shell of radius 0.735 m is of magnitude 868 N/C and points radially toward the center of the sphere.

(a) What is the net charge within the sphere's surface?
 nC 

(b) What is the distribution of the charge inside the spherical shell?
    

 

asked Feb 10, 2012 in Physics by apush ~Expert~ (552 points)
    

1 Answer

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Best answer

The net charge within the sphere's surface can be calculated using Gauss's Law, which states that

electric flux = E * A (*where E=electric field, A=area)

which is also equivalent to,

electric flux = (charge in) / (permittivity of free space)

Setting the two equations equal to one another will allow us to solve for Q.

 E * A = (charge in) / (permittivity of free space)

and therefore:

charge in =  E * A * (permittivity of free space)

given E = -868 N/C  (*its negative b/c it is pointing inward)

A = pi*r^2 , where r = 735m, and so A = pi*(735m)^2

permittivity of free space = 8.8542*10^-12 C^2/N*m^2

plugging all this back into the equation we get that:

 

charge in =  (-868 N/C) *(pi*(735m)^2) * ( 8.8542*10^-12 C^2/N*m^2) 

which simplifies to:

charge in =  -5.2174*10^-8 C  =  -52.174 nC

 

Part B:

The negative charge has a spherically symmetric charge distribution.

answered Feb 16, 2012 by awesome ~Expert~ (1,479 points)
selected Feb 18, 2012 by apush

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