Example of Electrostatics | Learn Physics

Example of Electrostatics

Question of Electrostatics:  Two fixed, equal positive charges each of magnitude 5 × 10^-5 C are located at point A and B separated by a distance of 6 m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of line AB. When the moving charge reaches the pt. C at a distance of 4 m from O, it has a K.E. of 4 J. calculate the distance of the farthest point D at which the negative charge will reach before returning to C.

Solution:

Here q = 5 × 10^-5 C

AB = 6 m, OC=4 m

BC =AC = √(3² + 4²)= 5 m

P.E. at C=[q(-q)+q(-q)]/4πϵ₀AC

=-9 joule

Total energy at C = P.E. +K.E. = +4-9=-5J

If r is the distance AD = BD, then

P.E. at D = [-2(9 × 10⁹)( 5 × 10^-5)²]/r=-45/r

As charge stops at D, K.E. at D =0

Total energy at D = K.E. + P.E. = 0-45/r=-45/r J

As total energy of the system remains constant,

-45/r =-5   or r= 9 m

Now OD =√(AD²-OA²) = √(9²-3²) = √72 = 6√2 m