Saved Bookmarks
| 1. |
There are three concentric metallic spherical shells of radii R, 2R and 3q, respectively. The middle spherical shell is connected to earth. Let q_(1) be the charge appearing on the inner surface of the middle sphere and q_(2) be the charge on outer surface of the middle sphere. Similarly q_(3) is the charge appearing on inner surface of outermost sphere and q_(4) is the charge appearing on outer surface of outermost sphere. |
|
Answer» `{:(P,Q,R,S),(4,3,1,2):}`  Since electric field intensity inside the innermost sphere is zero so no charge can stay on the inner surface and all charge q of inner sphere will stay on its outer surface only as shown in the figure. Metal faces parallel to each other must carry equal and opposite charges and hence charge appearing on inner surface of middle sphere must be -q. Let charge on outer surface of middle sphere be q. so that its electric potential may become zero. We shall calculate q. later by equating potential of middle sphere to zero. For now charge -q. must appear on inner surface of the outermost sphere, which is isolated. Hence, the net charge on it must remain 3q and hence the charge on outer surface must be `3q+q.`. Now we can calculate the value of q. by equating potential of middle sphere equal to zero because this sphere is connected to earth and the conductor connected to earth always ACQUIRE zero potential. `V_("middle-sphere")=(1)/(4pi epsilon_(0))[(q)/(2R)+(q.-q)/(2R)+(3q)/(3R)]` Note that we have used only the net charge on spheres to calculate potential of the middle sphere. We know that middle sphere is earthed hence: `V_("middle-sphere")=(1)/(4pi epsilon_(0))[(q)/(2R)+(q.-q)/(2R)+(3q)/(3R)]=0 rArr q.=-2q` Now we can WRITE the charges appearing on different faces: `q_(1)= -q` `q_(2)=-2q` `q_(3)=2q` `q_(4)=q` We can see that option (a) is correct. |
|