Saved Bookmarks
| 1. |
Small mercury drops of the same size are charged to the same potential V. If n such drops coalesce to form a single large drop, then calculate its potential. |
|
Answer» Solution :Let r be the radius of a small DROP and R that of the large drop. Then, since the volume remains conserved, `(4)/(3) pi R^(2)= (4)/(3) pi r^(2) n "" rArrR^(3) = r^(3) n ` R = `r^(3) (n)^(1//3)` Further, since the total CHARGE remains conserved, we have using Q = CV `C_("large") V = n C_("small"^(V))` Where V is the potential of the large drop. `4 pi epsilon_(0) RV = n (4 pi epsilon_(0) r)_(v)` V= `("nrv")/(R) = ("nrv")/(r(n)^(1//3))` `V = vn^(2//3)` |
|