Two fixed, equal, positive charges, each of magnitude `5xx 10^-5` C are located at points A and B separated by a distance of 6 m. An equal and opposite charge moves towards them along the fine COD, the perpendicular bisector of the line AB. The moving charge when it reaches the Point C at a distance of 4 m from 0, has a kinetic energy of 4 J. Calculate the distance of the farthest point D which the negative charge will reach before returning towards C.

Answer

8.48M

Explanation

Equating the energy of (-q) at C and D
`K_C+U_C=K_D+U_D`
`K_C=4J , U_C=2[1/(4pi epsilon_0)((q)(_q))/(AC)]=(-2xx9xx10^9xx(5xx10^(-5))^2)/(AD)=-9J`
`K_D=0 and U_D=2[1/(4pi epsilon_0)((q)(-q))/(AD)=(-2xx9xx10^2xx(5xx10^(-5))^2)/(AD)=(45)/(AD)`
Substituting these values in Eq.(i)
`4-9=0-(45)/(AD) :. AD=9m :. OD=sqrt(AD^2-OA^2)=sqrt((9)^2-(3)^2)=sqrt(81-9)=8.48m`