The dome of a Van de Graaff generator receives a charge of 0.00011 C. The radius of the dome is 5.2 m. Find the strength of the electric field inside the dome. The Coulomb constant is 8.98755 × 109 N m2 /C 2 , and the acceleration due to gravity is 9.8 m/s 2 . 1. 14062.2 N/C 2. −36561.8 N/C 3. 36561.8 N/C 4. 4.0218 N/C 5. 0 N/C 6. 0.386711 N/C 012 (part 2 of 3) 10.0 points Find the strength of the electric field at the surface of the dome. Answer in units of N/C. 013 (part 3 of 3) 10.0 points Find the strength of the electric field at a distance of 12.6 m from the center of the dome. Answer in units of N/C

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Kazeemsodikisola Kazeemsodikisola · Answer 1 · 2020-02-27T10:46:44+01:00

Answer:

Explanation:

Given that

K=8.98755×10^9Nm²/C²

Q=0.00011C

Radius of the sphere = 5.2m

g=9.8m/s²

1. The electric field inside a conductor is zero

εΦ=qenc

εEA=qenc

net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero

This surface encloses no charge, and thus qenc=0. Gauss’ law.

Since it is inside the conductor

E=0N/C

2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as

F=kq1/r²

F=kQ/r²

F=8.98755E9×0.00011/5.2²

F=36561.78N/C

The electric field at the surface of the conductor is 36561N/C

Since the charge is positive the it is outward field

3. Given that a test charge is at 12.6m away,

Then Electric field is given as,

E=kQ/r²

E=8.98755E9 ×0.00011/12.6²

E=6227.34N/C