Charge of uniform density (40 pC/m2) is distributed on a spherical surface (radius = 1.0 cm), and a second concentric spherical surface (radius = 3.0 cm) carries a uniform charge density of 60 pC/m2. What is the magnitude of the electric field at a point 4.0 cm from the center of the two surfaces?

4.1 N/C

Explanation:

First of all, we know from maths that the surface area of a sphere = 4πr²

Charge on inner sphere ..

Q(i) = 40.0*10^-12C/m² x 4π(0.01m)²

Q(i) = 5.03*10^-14 C

Charge on outer sphere

Q(o) = 60*10^-12 x 4π(0.03m)²

Q(o) = 6.79*10^-13 C

Inner sphere has a - 5.03*10^-14C charge (-Qi) on inside of the outer shell. As a result, there is a net zero charge within the outer shell.

For the outer shell to show a NET charge +6.79*10^-13C, it's must have a +ve charge

= +6.79*10^-13C + (+5.03*10^-14C)

= +7.29*10^-13 C

Now again, we have

E = kQ /r²

E = (9.0*10^9)(+7.29*10^-13 C) / (0.04)²

E = 6.561*10^-3 / 1.6*10^-3

E = 4.10 N/C

Thus, the magnitude of the electric field is 4.1 N/C

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2