A 4.7 cm radius spherical conducting shell has a uniform surface charge density.What is the density if the electric field 1.5 m from its center is 17 N/C outward? A) 1.5 $\times$ 10^-7 C/m² B) 7.6 $\times$ 10^-6 C/m² C) 1.1 $\times$ 10^-7 C/m² D) 1.5 $\times$ 10^-9 C/m² E) 6.1 $\times$ 10^-7 C/m²

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Quizplus · Accepted Answer

The electric field outside a conducting shell is the same as if all the charge were concentrated at the center of the shell. So, we can use the formula for the electric field of a point charge:$$E = k\frac{q}{r^2}$$where:* E is the electric field* k is Coulomb's constant* q is the charge* r is the distance from the chargeWe are given that the electric field 1.5 m from the center of the shell is 17 N/C outward. So, we can plug these values into the formula and solve for q:$$17 N/C = (9 imes 10^9 N m^2/C^2)\frac{q}{(1.5 m)^2}$$$$q = 3.5 imes 10^{-9} C$$Now, we can use the formula for the surface charge density of a spherical conducting shell:$$\sigma = \frac{q}{4\pi r^2}$$where:* σ is the surface charge density* q is the charge* r is the radius of the shellWe are given that the radius of the shell is 4.7 cm = 0.047 m. So, we can plug these values into the formula and solve for σ:$$\sigma = \frac{3.5 imes 10^{-9} C}{4\pi (0.047 m)^2}$$$$\sigma = 1.5 imes 10^{-7} C/m^2$$Therefore, the surface charge density of the shell is 1.5 x 10^-7 C/m².

A 47 Cm Radius Spherical Conducting Shell Has a Uniform Surface