Saved Bookmarks
| 1. |
Consider two point charges q_(1) and q_(2) at rest as shown in the figure. They are separated by a distance of 1 m. Calculate the force experienced by the two charges for the following cases: (a) q_(1)=+2 mu C and q_(2)=+3mu C (b) q_(1)=+2muC and q_(2)=-3mu C (c) q_(1)=+2muCand q_(2)-3mu C kept in water (epsilon_(r)=80) |
Answer» Solution : (a) `q_(1)=+2 mu` C and `q_(2)=+3muC` and r=1 m. Both are positive charges . So the force will be repulsive Force EXPERIENCED by the charge `q_(2)` due to `q_(1)` is given by `vecF=(1)/(4piepsilon_(0))(q_(1)q_(2))/(r^(2))r_(12)` Here `r_(12)` is the unit vector from `q_(1)` to `q_(2)` . Since `q_(2)` is located on the right of `q_(1)` we have `hatr_(12)=HATI` , so that `vecF_(n)=(9xx10^(9)xx2xx10^(-6)xx3xx10^(-6))/(1^(2))hati[(1)/(4piepsilon_(0))=9xx10^(9)]` `= 54xx10^(-3)N hati` According to Newton .s third law the force experienced by the charge `q_(1)` due to `q_(2)` is `vecF_(12)=-vecF_(21)` So that `vecF_(12)=-54xx10^(-3)Nhati` The directions of `vecF_(21)` and `vecF_(12)` are shown in the above figure in CASE (a) (b) `q_(1)=+2muC, q_(2)=-3muC` and r=1 m. They are unlike charges . So the force will be attractive force experienced by the charge `q_(2)` due to `q_(1)` is given by `vecF_(21)=(9xx10^(9)XX(2xx10^(-6))xx(-3xx10^(-6))/(1^(2))hatr_(12)=-54xx10^(-3)N hati("Using " hatr_(12)=hati)` The charge `q_(2)` will experience an attractive force towards `q_(1)` which is in the negative x direction . According to Newton.s third law the force experienced by the charge `q_(1)` due to `q_(2)` is `vecF_(12)=-vecF_(21)` So that `vecF_(12)=54xx10^(-3)Nhati)` The directions of `vecF_(21)` and `vecF_(12)` are shown in the figure (case (b)). (c ) If these two charges are kept INSIDE the water then the forrce experienced by `q_(2)` due to `q_(1)vecF_(21)^(w)=(1)/(4piepsilon)(q_(1)q_(2))/(r^(2))hatr_(12)""` Since `epsilon=epsilon_(r)epsilon_(0)` , we have `vecF_(21)^(w) =(1)/(4piepsilon_(r)epsilon_(0))(q_(1)q_(2))/(r^(2)) hatr_(12)=(vecF_(21))/(epsilon_(r))` Therefore `hatF_(21)^(w)=(-54xx10^(-3)N)/(80)hati=-0.675xx10^(-3)Nhati` |
|