Question

A solid metal sphere with radius 0.430 m carries a net charge of 0.270 nC . Part A Find the magnitude of the electric field at a point 0.106 m outside the surface of the sphere. Express your answer using three significant figures. E

Answer 1

Answer:

8.46 N/C

Explanation:

Using Gauss law

$E=\frac {kQ}{r^{2}}$

Gauss's Law states that the electric flux through a surface is proportional to the net charge in the surface, and that the electric field E of a point charge Q at a distance r from the charge

Here, K is Coulomb's constant whose value is $9\times 10^{9} Nm^{2}/C^{2}$

r = 0.43 + 0.106 = 0.536 m

$E=\frac {9\times 10^{9}\times 0.270\times 10^{-9}}{0.536^{2}}=8.4581755402094007\approx 8.46 N/C$