The figure shows a hollow conducting spherical shell with an inner radius of 8 cm and an outer radius of 10 cm. A charge, q, placed at the center of the shell causes an electric field at the inner surface of the shell, Ei𝐸𝑖 , which has a magnitude of 80 N/C and points toward the center of the sphere. The electric field at the outer surface, Eo𝐸𝑜 , has a magnitude of 80 N/C and points away from the center of the sphere. Determine the charge of the particle, q. rev: 10_23_2013_QC_2344Multiple Choice+ 1.46 × 10−10 C+ 1.46 × 10-10 C0 C− 5.69 × 10−11 C- 5.69 × 10-11 C− 1.46 × 10−10 C- 1.46 × 10-10 C+ 5.69 × 10−11 C+ 5.69 × 10-11 C− 3.20 × 10−11 C- 3.20 × 10-11 C+ 8.89 × 10−11 C+ 8.89 × 10-11 C+ 3.20 × 10−11 C+ 3.20 × 10-11 C− 8.89 × 10−11 C
Question
The figure shows a hollow conducting spherical shell with an inner radius of 8 cm and an outer radius of 10 cm. A charge, q, placed at the center of the shell causes an electric field at the inner surface of the shell, Ei𝐸𝑖 , which has a magnitude of 80 N/C and points toward the center of the sphere. The electric field at the outer surface, Eo𝐸𝑜 , has a magnitude of 80 N/C and points away from the center of the sphere. Determine the charge of the particle, q. rev: 10_23_2013_QC_2344Multiple Choice+ 1.46 × 10−10 C+ 1.46 × 10-10 C0 C− 5.69 × 10−11 C- 5.69 × 10-11 C− 1.46 × 10−10 C- 1.46 × 10-10 C+ 5.69 × 10−11 C+ 5.69 × 10-11 C− 3.20 × 10−11 C- 3.20 × 10-11 C+ 8.89 × 10−11 C+ 8.89 × 10-11 C+ 3.20 × 10−11 C+ 3.20 × 10-11 C− 8.89 × 10−11 C
Solution
The electric field E due to a point charge q is given by Coulomb's law:
E = k*q/r^2
where k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge.
Given that the electric field E at the inner surface of the shell is 80 N/C, we can rearrange the equation to solve for q:
q = E*r^2/k
Substituting the given values (E = 80 N/C, r = 8 cm = 0.08 m, k = 8.99 x 10^9 N m^2/C^2), we get:
q = 80 N/C * (0.08 m)^2 / (8.99 x 10^9 N m^2/C^2) = 5.69 x 10^-12 C
This charge is the charge induced on the inner surface of the shell by the charge at the center.
Since the shell is neutral, the charge on the outer surface must be equal and opposite to the charge on the inner surface. Therefore, the charge of the particle at the center, which caused this distribution of charge on the shell, must be -5.69 x 10^-12 C.
So, the correct answer is -5.69 x 10^-11 C.
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