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

`Va-Vb=168KV`

Step-by-step 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=
`√(0.3^2+0.4^2) =0.5`

at opposite sides

Mathematically Va can represented as

`Va =k((4*10^_-_6)/(0.3) +(-2*10^_-_6)/(0.5) )`

`Va =9*10^9((4*10^_-_6)/(0.3) +(-2*10^_-_6)/(0.5) )`

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

`Va =84000V`

`Va =84KV`

Mathematically Vb is represented as

`Va =k((-4*10^_-_6)/(0.3) +(2*10^_-_6)/(0.5) )`

`Va =9*10^9((-4*10^_-_6)/(0.3) +(+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`