Question

An electron is located on the x axis at x0 = -9.31×10-6 m. Find the magnitude and direction of the electric field at x = 9.39 ×10-6 m on the x axis due to this electron. What is the magnitude in N/C? Is the direction positive x direction, negative x direction, or neither.

Answer 1

The magnitude of electric field is 4.08 N/C and direction of electric field is  negative x axis.

The electric field is given as,

                             $E=k\frac{q}{r^{2} } ,k=9*10^{9} Nm^{2}/C^{2}$

Where q is charge and r is distance.

Given that, $q=1.6*10^{-19} C,r=2*9.39*10^{-6} =18.78*10^{-6} m$

Substitute values in above equation.

             $E=9*10^{9}*\frac{1.6*10^{-19} }{(18.78*10^{-6} )^{2} } \\\\E=4.08 N/C$

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