A) increases K-1 times
B) increases K times
C) decreases K times
D) remains constant
Correct Answer: C
Solution :
Key Idea: For all dielectrics the value of relative permittivity is greater than one. From Coulomb's law the force of attraction between two charges \[{{\text{q}}_{1}}\]and \[{{\text{q}}_{2}}\]separated by a distance r is given by \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]where \[{{\varepsilon }_{0}}\]is the absolute permittivity of the material medium. For air\[\varepsilon ={{\varepsilon }_{0}}\] and for all other materials. \[\therefore \] \[\varepsilon =K\varepsilon & 0\] where K is dimensionless constant called dielectric constant or relative permittivity of the material. \[\therefore \] \[F=\frac{1}{4\pi {{\varepsilon }_{0}}K}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\] For all dielectrics the value of K is greater than 1, thus, if some dielectric is there between the charges, then the force between them decreases K times.
You need to login to perform this action.
You will be redirected in
3 sec
