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A point charge with charge q1 = 4.00 μC is held stationary at the origin. A second point charge with charge q2 = -4.20 μC moves from the point ( 0.135 m , 0) to the point ( 0.250 m , 0.250 m ). How much work W is done by the electric force on the moving point charge? Express your answer in joules. Use k = 8.99×109 N⋅m2/C2 for Coulomb's constant: k=14πϵ0. View Available Hint(s)

Answer :

okpalawalter8

Answer:

The value is

Explanation:

From the question we are told that

    The first charge is  [tex]q_1 = 4.00 \mu C = 4.00 *10^{-6} \ C[/tex]

   The second  charge is  [tex]q_2 = -4.20 \mu C = - 4.20 *10^{-6} \ C[/tex]

    The position of the first charge is  [tex](x , y) = (0\ m,0\ m )[/tex]

    The initial position of the second charge is  [tex](x , y) = (0.135 \ m , \ 0 )[/tex]

    The final  position of the second charge is  [tex](x , y) = (0.250 \ m , \ 0.250 \ m )[/tex]

Generally the  electric potential  energy of the second charge at initial position is mathematically represented as

       [tex]E_1 = \frac{k * q_1 * q_2 }{ r}[/tex]

Here  [tex]r= 0.135 \ m[/tex]

So

         [tex]E_1 = \frac{8.99 *10^{9} * 4.00 *10^{-6} * (-4.20 *10^{-6}) }{ 0.135}[/tex]

=>     [tex]E_1 = -1.119 \ J[/tex]

Generally the  electric potential  energy of the second charge at initial position is mathematically represented as

       [tex]E_2 = \frac{k * q_1 * q_2 }{ r}[/tex]

Here  [tex]r= \sqrt{0.250^2 + 0.250^2}[/tex]

 =>  [tex]r= 0.3536[/tex]

So

         [tex]E_2 = \frac{8.99 *10^{9} * 4.00 *10^{-6} * (-4.20 *10^{-6}) }{ 0.3536}[/tex]

=>     [tex]E_2 = - 0.4271 \ J[/tex]

Generally the workdone is mathematically represented as

     [tex]W = E_2 - E_1[/tex]

=>   [tex]W = -0.4271 - (-1.119 )[/tex]

=>   [tex]W = 0.6919 \ J[/tex]      

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