Answer:
[tex]Va-Vb=168KV[/tex]
Explanation:
From the question we are told that
Two points charge of 4\mu C and 2\mu C
Generally we find the Va and Vb individually to find there difference
Given a rectangle with two equal sides each,Assume lengths for bot sides
Length L=0.3
Breath B=0.4
Diagonal D=[tex]\sqrt{0.3^2+0.4^2} =0.5[/tex]
at opposite sides
Mathematically Va can represented as
[tex]Va =k(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )[/tex]
[tex]Va =9*10^9(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )[/tex]
[tex]Va =9*10^9(0.00001333333-0.000004} )[/tex]
[tex]Va =84000V[/tex]
[tex]Va =84KV[/tex]
Mathematically Vb is represented as
[tex]Va =k(\frac{-4*10^_-_6}{0.3} +\frac{2*10^_-_6}{0.5} )[/tex]
[tex]Va =9*10^9(\frac{-4*10^_-_6}{0.3} +\frac{+2*10^_-_6}{0.5} )[/tex]
[tex]Va =9*10^9(-0.00001333333+0.000004} )[/tex]
[tex]Va =-84000V[/tex]
[tex]Va =-84KV[/tex]
Therefore
[tex]Va-Vb=84-(-84)\\Va-Vb=84+84\\Va-Vb=168KV[/tex]
