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A point charge q1 = 4.10 nC is placed at the origin, and a second point charge q2 = -2.90 nC is placed on the x-axis at x = + 20.0 cm. A third point charge q3 = 2.00 nC is to be placed on the x-axis between q1 and q2. (Take as zero the potential energy of the three charges when they are infinitely far apart.)
1. What is the potential energy of the system of the three charges if q3 is placed at x = + 11.0 cm?2. Where should q3 be placed between q1 and q2 to make the potential energy of the system equal to zero?

Respuesta :

Answer:

1)  U = -4.436 10-7 J ,  2)   r13 = 0.6937 m

Explanation:

The electric potential energy is given by the equation

      U = k Σ [tex]q_{i}[/tex] [tex]q_{j}[/tex]  / [tex]r_{ij}[/tex]

1. Let's apply this equation to our case

      U = k (q1q3 / r13 + q1 q2 / r12 + q2 q3 / r23)

Let's reduce all magnitudes to the SI system

    q1 = 4.10 nC = 4.10 10-9 C

    q2 = -2.90 nC = -2.90 10-09 C

    q3 = 2.00 nC = 2.00 10-9 C

Let's look for the distances

    r13 = 11-0 = 11 cm = 11 10-2 m

    r12 = 20-0 = 20 cm = 20 10-2 m

    r23 = 20 -11 = 9 cm = 9.0 10-2 m

Let's calculate the electric potential

     U = 8.99 109 [(4.10 2.00) 10-18 / 11 10-2 + ​​(4.10 (-2.90)) 10-18 / 20 10-2 + ​​(-2.90 2.00) 10-18 / 9.0 10-2]

     U = 8.99 [0.7455 - 0.5945 - 0.6444) 10-7

     U = -4.436 10-7 J

2.  ask to find the position of q3 for the energy to be zero

    U = 0 = k (q1q3 / r13 + q1 q2 / r12 + q2 q3 / r23)

In this case the distance between 1 and 2 is fixed, since the load that is placed is 3

     q1 q2 / r12 = - q1 q3 / r13 - q2 q3 / r23

     -11.89 10-18 / 20 10-2 = -8.20 10-18 / r13 + 5.810-18 / r23

     -0.5945 102 = -8.20 / r13 + 5.810 / r23

Let's relate the distances, the maximum separation esr12 which is 20 10-2 m

     r12 = r13 + r23

     r13 = r12 -r23

    r13 = 2010-2 - r23

we replace

    -0.5945 102 = -8.20 / (20 10-2 - r23) + 5.810 / r23

Let's solve this equation

    -0.5945 102 (2010-2 - r23) = -8.20 r23 + 5.810 (2010-2 - r23)

    - 11.89 + 0.5945 r23 = -8.20 r23 + 1,162 - 5,810 r23

     -11.89 - 1,162 = - 8.20 r23 -5,810 r23 - 0.5945 r23

     -13.052 = r23 (-14.6045)

     r23 = 13.052 / 14.6045

    r23 = 0.8937 m

The distance from the origin is

    r13 = -r12 + r23

    r13 = -20 10-2 + ​​0.8937

    r13 = 0.6937 m

onyebuchinnaji

(a) The potential energy of the system of the three charges is 0.9 x 10⁻⁷ J.

(b) The position of charge 3 to make the Potential energy of the system equal to zero is 11.7 cm.

Total potential energy of the system

The total potential energy of the system is calculated as follows;

[tex]U = qV = V_{13}q_1 + V_{23}q_2 \\\\U= \frac{kq_3}{r_1_3}q_1 + \frac{kq_3}{r_2_3}q_2 \\\\U = \frac{9\times 10^9 \times 2 \times 10^{-9} }{0.11}(4.1 \times 10^{-9}) + \frac{9\times 10^9 \times 2\times 10^{-9}}{0.2 - 0.11} (-2.9 \times 10^{-9})\\\\U = 6.7\times 10^{-7} \ J \ - 5.8 \times 10^{-7} \ J\\\\U = 0.9 \times 10^{-7} \ J[/tex]

Position for zero potential energy

The position of charge 3 to make the Potential energy of the system equal to zero is determined as follows;

let the position of charge 3 = x

[tex]\frac{kq_3}{r_1_3}q_1 + \frac{kq_3}{r_2_3}q_2 = 0 \\\\\frac{9\times 10^9 \times 2 \times 10^{-9} }{x-0}(4.1 \times 10^{-9}) + \frac{9\times 10^9 \times 2\times 10^{-9}}{0.2 - x} (-2.9 \times 10^{-9}) = 0\\\\\frac{7.38 \times 10^{-8}}{x} - \frac{5.22 \times 10^{-8}}{0.2 - x} = 0\\\\5.22 \times 10^{-8} x = 7.38 \times 10^{-8} (0.2 - x)\\\\5.22x = 7.38(0.2 - x)\\\\5.22x = 1.476 - 7.38x\\\\12.6x = 1.476\\\\x = 0.117 \ m\\\\x = 11.7 \ cm[/tex]

Thus, the position of charge 3 to make the Potential energy of the system equal to zero is 11.7 cm.

Learn more about potential energy here: https://brainly.com/question/1242059

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