Question

A spherical conductor of radius = 1.5 cm with a charge of 3.9 pC is within a concentric hollow spherical conductor of inner radius = 3 cm, and outer radius = 4 cm, which has a total charge of 0 pC. What is the magnitude of the electric field 2.3 cm from the center of these conductors?

Answer 1

The answer is "66.351 N/C"

Step-by-step explanation:

Given:

`a=1.5\ cm= 1.5 * 10^(-2)\ m\\\\q_1=3.9\ pc\\\\b=3\ cm\\\\c= 4\ cm\\\\q_2=0 \ pc\\\\`

Using Gauss Law:

`\oint \vec{E} \cdot \vex{dA}= (Q_(enc))/(\varepsilon_0 )`

`E * 4 \pi\ r^2=(Q_(enc))/(\varepsilon_0)\\\\E= (Q_(enc))/(4 \pi\ r^2 \varepsilon_0)= (1)/(4 \pi \varepsilon_0) (Q_(enc))/(r^2)= (k_e\ Q_(enc))/(r^2)\\\\`

`=(9* 10^(9) * 3.9 * 10^(-12))/((2.3* 10^(-2))^2)\\\\=(35.1* 10^(-3)\ )/((2.3* 10^(-2))^2)\\\\=(35.1* 10^(-3)\ )/(5.29 * 10^(-4))\\\\=(35.1* 10 )/(5.29 )\\\\=(351)/(5.29 )\\\\=66.351\ (N)/(C)`