3.32/10 points I Previous Answers SerPSE9 25 P059 The electric potential immedia

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3.32/10 points I Previous Answers SerPSE9 25 P059 The electric potential immediately outside a charged conducting sphere is 190 V, and 10.0 cm farther from the center of the sphere the potential is 140 V. My Notes Ask Your (a) Determine the radius of the sphere 28 Cm (b) Determine the charge on the sphere 5.91 nC The electric potential immediately outside another charged conducting sphere is 210 V, and 10.0 cm farther from the center the magnitude of the electric field is 410 V/m (c) Determine all possible values for the radius of the sphere. (Enter your answers from smallest to largest. If only one value exists, enter "NONE" in the second answer blank.) 27.6 Your response differs from the correct answer by more than 100%. cm The correct answer is not zero. cm (d) Determine the charge on the sphere for each value of r. (If only one value exists, enter "NONE" in the second answer blank.) 345 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. nc 0.06 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. nC Need Help? Read Submit Answer -Save Progress Practice Another Version

Explanation / Answer

c)

let the radius of the sphere be "R"

V = electric potential just outside the sphere = 210 Volts

Q = charge on the sphere

Electric potential is given as

V = kQ/R

kQ = VR eq-1

r = distance from center where electric field is given = R + 0.1

E = magnitude of the electric field = 410 V/m

Electric field is given as

E = kQ/r2

using eq-1

E = VR/r2

E = VR/(R + 0.1)2

inserting the values

410 = (210)R /(R + 0.1)2

R = 0.036 m = 3.6 cm

and

R = 0.276 m = 27.6 cm

d)
For R = 0.036

using eq-1

kQ = VR

(9 x 109) Q = (210) (0.036)

Q = 8.4 x 10-10 C = 0.84 nC

For R = 0.276

using eq-1

kQ = VR

(9 x 109) Q = (210) (0.276)

Q = 6.44 x 10-9 C = 6.44 nC

Q = 8.4 x 10-10 C

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