Calculate the magnitude of the electric field at a point outside the sphere. Con

Question

Calculate the magnitude of the electric field at a point outside the sphere.

Conceptualize: Note how this problem differs from our previous discussion of Gauss's law. Now we are considering the electric field due to a distribution of charge. We found the field for various distributions of charge in the chapter entitled Electric Fields by integrating over the distribution. In this chapter, we find the electric field using Gauss's law.

Categorize: Because the charge is distributed uniformly throughout the sphere, the charge distribution has spherical symmetry and we can apply Gauss's law to find the electric field.

Analyze: To reflect the spherical symmetry, let's choose a spherical gaussian surface of radius r, concentric with the sphere, as shown in Figure (a). For this choice,

are parallel everywhere on the surface and

B/

Find the magnitude of the electric field at a point inside the sphere.

Analyze: In this case, let's choose a spherical gaussian surface having radius r < a, concentric with the insulating sphere (see Figure (b)). Let V' be the volume of this smaller sphere. To apply Gauss's law in this situation, recognize that the charge qin within the gaussian surface of volume V' is less than Q.

Calculate qin by using qin = V':

Apply Gauss's law in the region r < a:

Explanation / Answer

assumption:

charge on the sphere=Q

radius of the sphere=a

then for distance from center r>a:

charge enclosed by resulting gaussian surface=Q

if electric field at.r is given by E,

writing Gauss’ law:

epsilon*E*4*pi*r^2=charge enclosed=Q

==>E=Q/(4*pi*epsilon*r^2)

where epsilon=electrical permitivity of free space

example:

as field is downward, charge is negative.

distance=r=6371*1000 m

using the formula obtained earlier,

1.32*10^2=8.99*10^9*charge/(6371*10^3)^2

==>charge=1.32*10^2*(6371*1000)^2/(8.99*10^9)=5.9598*10^5 C

using the correct sign,

Qearth=-5.9598*10^5 C

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