A thin metallic spherical shell of radius .5m has a total charge of 8 μC uniformly distributed on it. At the center of the shell is placed a point charge of 4 μC. What is the electric field at a distance of .15 m from the center of the spherical shell?
I know that EA=q/e but I don't know how to use it in this context.
To find the electric field at a distance of 0.15 m from the center of the spherical shell, we can use the concept of superposition of electric fields. Since the spherical shell has a uniform charge distribution, the electric field due to the shell itself will be zero at all points inside the shell. Therefore, we only need to consider the electric field due to the point charge at the center.
Let's label the point charge at the center as Q1 = 4 μC and the charge on the spherical shell as Q2 = 8 μC.
Now, let's determine the electric field due to each of these charges separately.
1. Electric field due to the point charge at the center (Q1):
The electric field due to a point charge can be calculated using Coulomb's law:
E1 = k * (Q1) / r^2,
where k is the electrostatic constant (9 × 10^9 Nm^2/C^2) and r is the distance from the point charge.
In this case, since we're interested in the electric field at a distance of 0.15 m from the center of the shell, r = 0.15 m.
Plugging in the values, we get:
E1 = (9 × 10^9 Nm^2/C^2) * (4 × 10^-6 C) / (0.15 m)^2.
2. Electric field due to the charged spherical shell (Q2):
Since we are interested in the electric field at a distance inside the shell, the shell can be treated as a uniformly charged thin shell. The electric field inside a uniformly charged thin shell is zero.
Now, to find the net electric field at the given point, we'll add the electric fields due to each charge vectorially.
E_net = E1 + E2,
where E2 = 0 because of the uniformly charged thin shell's electric field being zero.
Therefore, the electric field at a distance of 0.15 m from the center of the spherical shell is given by:
E_net = (9 × 10^9 Nm^2/C^2) * (4 × 10^-6 C) / (0.15 m)^2.
Calculating this expression will give you the value of the electric field at the given point.