A) 2F
B) \[\frac{F}{2}\]
C) \[\frac{F}{4}\]
D) \[\frac{F}{8}\]
Correct Answer: D
Solution :
Key Idea: Charge is placed end-on position of dipole. The electric field at a point distance r from electric dipole is \[E=\frac{1}{4\pi Eo}.\frac{p}{{{r}^{3}}}\] (end-on position) where p is dipole moment and r is the distance of charge from centre of dipole. or \[E\propto \frac{1}{{{r}^{3}}}\] Force on charge is \[F=QE\] or \[F\propto \frac{1}{{{r}^{3}}}\] \[\therefore \] \[\frac{{{F}_{2}}}{{{F}_{1}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}\] Given, \[{{r}_{1}}=r,{{r}_{2}}=2r,{{F}_{1}}=F\] \[\therefore \] \[\frac{{{F}_{2}}}{F}={{\left( \frac{r}{2r} \right)}^{3}}=\frac{1}{8}\] Or \[{{F}_{2}}=\frac{F}{8}\]
You need to login to perform this action.
You will be redirected in
3 sec
