Question

two equal point charges each of micro coulomb are separated by certain distance in metres.if they are located at (i+j+k) and (2i+3j+4k),then the electrostatic force between them is

Answer 1

F will be minimum when q1andq2q1andq2 are minimum i.e.q1=q2=1.6×10−19i.e.q1=q2=1.6×10−19
Also, given r=1mr=1m.
Fmin=9×109N×(1.6×10−10)2Fmin=9×109N×(1.6×10−10)2
Fmin=2.3×10−28N.

Answer 2

first of all we need distance between two point charges I. e R
so we have Given their position vector
let r1 be the position vector of 1 point charge is I + j + K
let r2 be position vector of 2nd point charge is 2i + 3j + 4k
so R will be r2 - r 1 = 1 i^ + 2 j + 3 j = 6
R ²= 36 m²
now charge will be IE. q²=
(9.11* 10 -19 c * 10 -6.)² ( micro columb )
Now apply formula
F = K q²/ R²
u will get an ans