Question

Two points charge of 4\mu C and 2\mu C are placed at theopposite corners of a rectangle. What is the potential difference Va- Vb

Answer 1

Answer:

$Va-Vb=168KV$

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=$\sqrt{0.3^2+0.4^2} =0.5$

at  opposite sides

Mathematically Va can represented as

$Va =k(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )$

$Va =9*10^9(\frac{4*10^_-_6}{0.3} +\frac{-2*10^_-_6}{0.5} )$

$Va =9*10^9(0.00001333333-0.000004} )$

$Va =84000V$

$Va =84KV$

Mathematically Vb is  represented as

$Va =k(\frac{-4*10^_-_6}{0.3} +\frac{2*10^_-_6}{0.5} )$

$Va =9*10^9(\frac{-4*10^_-_6}{0.3} +\frac{+2*10^_-_6}{0.5} )$

$Va =9*10^9(-0.00001333333+0.000004} )$

$Va =-84000V$

$Va =-84KV$

Therefore

$Va-Vb=84-(-84)\\Va-Vb=84+84\\Va-Vb=168KV$