Question

A thin ball shell is shaped like a hollow ball with a radius of 42.8 cm. The material is a conductor with charge 0.8 nC A point charge of -5.4 nC is placed in the center of the ball. How many N / C is the electric field just outside the ball? Remember positive is outwards I did this: got Total charge Qencloced = -4.6*10^-9 C Found the field E=(Qencloced/Epsilon) divided by 4*pi*speed squared Got 225.80 N/C. The answer in my (really annoying online test) is 265 N/C

Answer 1

Answer:

The Electric field will be 225.92 N/C

Explanation:

Given :

• Radius of the hollow sphere R=42.8 cm
• Initial charge on the conducting sphere is $Q_1=0.8\times10^{-9}\ \rm C$
• Magnitude of the point charge $q_2=-5.4\times10^{-9}\ \rm C$

We know that the flux of the electric field through a close surface cab be calculated by using Gauss Law which is

$\int EdA=\dfrac{Q_{in}}{\epsilon_0}\\\\E\times4\pi\times 0.428^2=\dfrac{(0.8-5.4)\times10^{-9}}{\epsilon_0}\\\\E=225.92\ \rm N/C$

Hence the Electric Field is calculated.

The answer founded out by you seems to be correct and very close to the exact answer and the concept you have used in your answer is also correct.