Saved Bookmarks
| 1. |
The charges of an electric dipole are respectively 32xx10^(-7) coulomband -32xx10^(-7) coulomb are separated by a distance of10 cm . Find the field at a point situated at a distance of 8 cm fromeach charge. |
Answer» Solution :Suppose `E_(1)` is the FIELD due to `+32xx10^(-7)` COULOMB charge and `E_(2)` is the field due to the `-32xx10^(-7)` coulomb charge at C, then  `E_(1)=(1)/(4piepsilon_(0))(q_(1))/(r^(2))` `=(9XX10^(9)xx32xx10^(-7))/((0.8)^(2))` `=4.5xx10^(6)N//C` . Along CP and `E_(2)=(9xx10^(9)xx32xx10^(-7))/((.08)^(2))` `=4.5xx10^(6)N//C`.Along CQ But from `Delta.s` PCR and CAB `(CR)/(AB)=(CP)/(CA)` or `(E)/(0.1)=(4.5xx10^(6))/(.08)` or , `E=(0.10xx4.5xx10^(6))/(.08)` `=5.625xx10^(6)` `E=5.625xx10^(6)("newton")/("coulomb")` |
|