Question

A hollow conducting sphere has an inner radius of r1 = 0.19 m and an outer radius of r2 = 0.34 m. The sphere has a net charge of Q = 1.8E-06 C. What is the field E1 in N/C 1 m from the sphere's outer surface?

Answer 1

Answer:

9022.0539 v/m

Explanation:

We know as given:

r1 = 0.19

r2 = 0.34

Q = 1.8 × 10⁻⁶ C

and it is known that E1 = k × q/r²    (k = 9× 10⁹)

= 9× 10⁹ ×  1.8 × 10⁻⁶/ (1+0.34)²

= 9022.0539 v/m

Thus, the correct answer is = E1 = 9022.0539 v/m.

Answer 2

We have that the electric field in N/C 1m from the sphere's outer surface is mathematically given as

E = 9022.0539 N/C

Electric field

Question Parameters:

• r1 = 0.19
• r2 = 0.34
• Q = 1.8 × 10⁻⁶ C

Generally the equation for the Electric field  is mathematically given as

E= k × q/r²    

Where

k = 9× 10^9

E= 9× 10^9 ×  1.8 × 10^{-6}/ (1+0.34)2

E= 9022.0539 N/C

Therefore

The electric field in N/C 1m from the sphere's outer surface

E = 9022.0539 N/C.

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