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?
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Ответ:
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
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