12. o/12 points | Previous Answers SerPSET9 25.P059. My Notes Ask Your Teacher T
ID: 3280326 • Letter: 1
Question
12. o/12 points | Previous Answers SerPSET9 25.P059. My Notes Ask Your Teacher The electric potential immediately outside a charged conducting sphere is 190 sphere the potential is 130 V and 10.0 cm farther from the center of the (a) Determine the radius of the sphere. 6842-0 x Your response differs from the correct answer by more than 10%. Double check your calculations. cm (b) Determine the charge on the sphere. 1.44 Your response differs from the correct answer by more than 10%. Double check your calculations. nc The electric potential immediately outside another charged conducting sphere is 220 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 ri cm 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.) q1 = q2 = nC nc Need Help?Read tExplanation / Answer
Given,
V1 = 190 V ; r = 10 cm
V2 = 130 V
V1 = kQ/R
V2 = kQ/(R + 0.1)
V2/V1 = R/(R + 0.1)
130/190 = R/(R + 0.1)
0.684 R + 0.068 = R
0.316 R = 0.068 => R = 0.215 m
Hence, R = 0.215 m = 21.5 cm
Q = V1 R/k
Q = 190 x 0.215/9 x 10^9 = 4.54 x 10^-9 C
Hence, Q = 4.54 x 10^-9 C
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Given,
V1 = 220 V ; E = 410 V/m
V2 = 410 x 0.1 = 41 V
V1 = kQ/R
V2 = kQ/(R + 0.1)
V2/V1 = R/R + 0.1
41/220 = R/(R + 0.1)
0.186 R + 0.019 = R
0.019 = 0.814R =>R = 0.0233 m = 23.3 cm
Hence, R1 = 2.33 cm ; R2 = 12.33 (2.33 +10 )
Q = V1 R/k
Q = 220 x 0.0233/9 x 10^9 = 0.57 x 10^-9 C
Hence, Q = 0.57 x 10^-9 C
for both the values same charge stays.
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