Saved Bookmarks
| 1. |
Two conducting spheres of radius r_(1) =8 cm and r_(2) =2 cm are separated by a distance much larger than 8 cm and are connected by a thin conducting wire as shown in the figure . A total charge of Q=+100 nC is placed on one of the spheres . After a fraction of a second the charge Q is redistributed and both the spheres attain electrostatic equilibrium . (a) Calculate the charge and surface charge density on each sphere. (b) Calculate the potential at the surface of each sphere . |
|
Answer» Solution :( a) The electrostatic potential on the surface of the sphere A is `V_(A)=(1)/(4piepsillon_(0))(q_(1))/(r_(1))` The electrostatic potential on the surface of the sphere A is `V_(B) = (1)/(4piepsilon_(0))(q_(2))/(r_(2))` Since `V_(A)=V_(B)`. We have `(q_(1))/(r_(1))=(q_(2))/(r_(2))implies q_(1)=((r_(1))/(r_(2)))q_(2)` But from the conservation of total charge Q`=q_(1)+q_(2)` we get `q_(1)=Q-q_(2).` By substituting this in the above equation , `Q-q_(2)=((r_(1))/(r_(2)))q_(2)" " "so that" " "q_(2)=Q((r_(2))/(r_(1)+r_(2)))` Therefore `q_(2)=100xx10^(-9)xx((2)/(10))=20nC` and `q_(1)=Q-q_(2)=80nC` The electric charge density for sphere A is `sigma_(1)=(q_(1))/(4pir_(1)^(2))` The electric charge density for sphere B is `sigma_(2)=(q_(2))/(4pir_(2)^(2))` Therefore`sigma_(1)=(80xx10^(-9))/(4xx64xx10^(-4))=0.99xx10^(-6)CM^(-2)` and `sigma_(2)=(20xx10^(-9))/(4pixx4xx10^(-4))=3.9XX10^(-6)cm^(-2)` Note that the surface charge density is greater on the smaller sphere compared to the largersphere `(sigma_(2)~~4sigma_(1)) ` which confirms the result `(sigma_(1))/(sigma_(2))=(r_(2))/(r_(1))` . The potential on both sphere is the same . So we can calculate the potential on any on of the spheres . `V_(A)=(1)/(4piepsilon_(0))(q_(1))/(r_(1))=(9xx10^(9)xx80xx10^(-9))/(8xx10^(-2))=9kV` |
|