Question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer 1

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The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

    The third charge is  $q_3 = 55 nC = 55 *10^{-9} C$

    The position of the third charge is  $x = -1.220 \ m$

     The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

     The position of the first charge is $x_1 = -1.650m$

      The second charge is  $q_2 = 32 nC = 32 *10^{-9} C$

      The position of the second charge is  $x_2 = 0 \ m$  

The distance between the first and the third charge is

      $d_{1-3} = -1.650 -(-1.220)$

     $d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is  

     $F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value  $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

      $F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

       $F_{1-3} = 4.28 *10^{-5} \ N$

 The distance between the second and the third charge is      

  $d_{2-3} = 0- (-1.22)$

   $d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

       $F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

       $F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

       $F_{2-3} = 1.06*10^{-5} N$

The net force is

      $F_{net} = F_{1-3} -F_{2-3}$

substituting values

    $F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

    $F_{net}= 3.22*10^{-5} \ J$

Answer 2

The net force on the third charge is -3.216 x 10⁻⁵ N.

The given parameters:

• Third charge, q3 = 55 nC
• First charge, q1 = -16 nC

• Second charge q2 = 32 nC
• Position of third charge, x3 = -1.22 m
• Position of first charge, x1 = -1.65 m
• Position of the second charge, x2 = 0 m

The force on the third charge due to the first charge is calculated as follows;

r = x1 - x3

r = -1.65 - (-1.22)

r = -0.43 m

$F_{13} = \frac{9\times 10^9 \times (-16 \times 10^{-9}) \times (55 \times 10^{-9})}{(0.43)^2} \\\\F_{13} = -4.28 \times 10^{-5} \ N$

The force on the third charge due to the second charge is calculated as follows;

$F_{23} = \frac{kq_2 q_3}{r^2} \\\\F_{23} = \frac{9\times 10^9 \times (32 \times 10^{-9}) \times (55 \times 10^{-9})}{(1.22)^2} \\\\F_{23} = 1.064 \times 10^{-5} \ N$

The net force on the third charge is calculated as follows;

$F_{net} = F_{13} + F_{23}\\\\F_{net} = -4.28 \times 10^{-5} \ N \ + \ 1.064 \times 10^{-5} \ N\\\\F_{net} = -3.216 \times 10^{-5} \ N$

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