Question

Q. Charges +2Q and -Q are placed as shown in figure . The point at which electric field intensity is zero will be

-Q +2Q
_______________________
$\stackrel{}{\leftarrow }$
E

a. Somewhere between -Q and +2Q
b. Somewhere on the left of -Q
c. Somewhere on the right of + 2Q
d. Somewhere on the right bisector of line joining -Q and +2Q

Answer 1

Dear Student,

$$\begin{gathered} Because,-Qhassmallermagnitude,Henceelectricfieldwillbezeroatapointcloseto-Qand \\ pointwillliesalongthelinejoingthechargesandleftsideofof-Q.Tofindthelocationofthispointletthedistacebetweenthechrgesberand \\ thedis\tan ceofthepointatwhichfieldiszerobexfrom-Q.Hencedis\tan ceofthepointfromthe+2Qwillber+x. \\ Togetzerofield \\ \frac{kQ}{x^2}=\frac{2kQ}{{\left(r+x\right)}^2} \\ \frac{1}{x^2}=\frac{2}{{\left(r+x\right)}^2} \\ \frac{1}{x}=\frac{\sqrt{2}}{\left(r+x\right)}\Rightarrow \boxed{x=\frac{r}{\sqrt{2}-1},Leftof-Q} \end{gathered}$$
Regards.