A 20 pC charge is placed at the origin and a -30 pC is places 3m to the right of it. What is the magnitude and direction of the electric field 7m meters from the origin?
Answer
Electric field due to any point charge is given
"E=\\frac{kq}{r^2}"
Electric field due 20 pC charge
"E_1=\\frac{9\\times10^9\\times20\\times10^{-12}}{7\\times7}(-i)"
"=3.7\\times10^{-3}(-i)" V/m
"E_2=\\frac{9\\times10^9\\times30\\times10^{-12}}{4\\times4}(i)"
= "=16.9\\times10^{-3}(i)" V/m
So resultant
The magnitude and direction of the electric field 7m meters from the origin
Is
E="E_1+E_2"
"=13.2\\times10^{-3}(i)" v/m
Direction in positive x direction.
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