Question #151760
The electric potential immediately outside a charged conducting sphere is 200 V, and 10.0 cm farther from the center of the sphere the potential is 150 V. Determine (a) the radius of the sphere and (b) the charge on it. 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 400 V/m. Determine (c) the radius of the sphere and (d) its charge on it. (e) Are the answers to parts (c) and (d) unique?
1
Expert's answer
2020-12-21T04:00:06-0500

a) ϕ1=kqr\phi_1=k\frac{q}{r} and ϕ2=kqr+0.1\phi_2=k\frac{q}{r+0.1} \to ϕ1r=ϕ2r+0.1ϕ2\phi_1r=\phi_2r+0.1\phi_2


r=0.1ϕ2ϕ1ϕ2=0.1150200150=0.3(m)=30(cm)r=\frac{0.1\phi_2}{\phi_1-\phi_2}=\frac{0.1\cdot150}{200-150}=0.3(m)=30(cm)


b) q=(r+0.1)ϕ2k=(0.3+0.1)1509109=6.67(nC)q=\frac{(r+0.1)\phi_2}{k}=\frac{(0.3+0.1)\cdot150}{9\cdot10^9}=6.67(nC)


c) 210=kqr210=k\frac{q}{r} and 400=kq(r+0.1)2400r2130r+4=0400=k\frac{q}{(r+0.1)^2}\to 400r^2-130r+4=0\to


r1=3.44(cm)r_1=3.44(cm) and r2=29.1(cm)r_2=29.1(cm) .


d) q1=210r1k=2100.003449109=803(pC)q_1=\frac{210\cdot r_1}{k}=\frac{210\cdot 0.00344}{9\cdot10^9}=803(pC)


q2=210r2k=2100.2919109=6.79(nC)q_2=\frac{210\cdot r_2}{k}=\frac{210\cdot 0.291}{9\cdot10^9}=6.79(nC)


e) No.Two answers exist for each part.





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