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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

Sagot :

okpalawalter8

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]

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