The electric potential immediately outside a charged conducting sphere is 230 V,
ID: 2077905 • Letter: T
Question
The electric potential immediately outside a charged conducting sphere is 230 V, and 10.0 cm farther from the center of the sphere the potential is 130 V. (a) Determine the radius of the sphere. cm (b) Determine the charge on the sphere. nC The electric potential immediately outside another charged conducting sphere is 260 V, and 10.0 cm farther from the center the magnitude of the electric field is 380 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.) r_1 = cm r_2 = 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.) q_1 = nC q_2 = nCExplanation / Answer
electric potential V1 = 230 V
electric potential V2 = 130 V
electric potential at the surface of the sphere is
V1 = kq/ R
230 = kq/ R
kq = 230*R ........... (1)
potential at (R + 10) from the center of the sphere is
V2 = kq/(R + 10)
130 = kq/(R + 10)
kq = 130(R + 10) ........... (2)
compare eq (1) and (2), we get
230*R = 130(R + 10)
radius of the sphere is
100 R = = 1300
R = 13 cm
R = 0.13 m
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substitute the value R = 13 cm in eq (1), we get
kq = 230*0.13
(9*109)(q) = 29.9
q = 3.32*10-9 C
q = 3.32 nC
charge on sphere is
q = 3.32 nC
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electric potential at the surface of the sphere is
V1 = kq/ R
260 = kq/ R
kq = 260*R ................... (3)
electric field at (R + 10) from the center of the sphere is
E = kq/(R + 10)2
380 = kq/(R + 10)2
kq = 380(R + 10)2 ..................... (4)
compare eq (3) and (4), we get
260*R = 380(R + 10 cm)2
260*R = 380(R + 0.1 m)2
380 R2 - 184 R + 3.8 = 0
solve the above quadratic equation , we get
R1 = 0.4626 m or R2 = 0.02162 m
R1 = ˜ 46.3 cm or R2 =˜ 2.16 cm
radius of the sphere is,
R = 46.3 cm (or) R = 2.16 cm
if R = 46.3 cm ,
substitute the value R = 46.3 cm in eq (3), we get
charge on the sphere q = 13.37 nC
if R = 2.16 cm ,
substitute the value R = 2.16 cm in eq (3), we get
charge on the sphere q = 624 pC
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