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Vera_Pavlovna [14]
2 years ago
9

Charge q1 = +2.00 μC is at -0.500 m along the x axis. Charge q2 = -2.00 μC is at 0.500 m along the x axis. Charge q3 = 2.00 μC i

s at 0.500 m along the y axis. What is the magnitude of electrical force on charge q3 due to the others?
Physics
1 answer:
Kobotan [32]2 years ago
8 0

The magnitude of <em>electrical</em> force on charge q_{3} due to the others is 0.102 newtons.

<h3>How to calculate the electrical force experimented on a particle</h3>

The vector <em>position</em> of each particle respect to origin are described below:

\vec r_{1} = (-0.500, 0)\,[m]

\vec r_{2} = (+0.500, 0)\,[m]

\vec r_{3} = (0, +0.500)\,[m]

Then, distances of the former two particles particles respect to the latter one are found now:

\vec r_{13} = (+0.500, +0.500)\,[m]

r_{13} = \sqrt{\vec r_{13}\,\bullet\,\vec r_{13}} = \sqrt{(0.500\,m)^{2}+(0.500\,m)^{2}}

r_{13} =\frac{\sqrt{2}}{2}\,m

\vec r_{23} = (-0.500, +0.500)\,[m]

r_{23} = \sqrt{\vec r_{23}\,\bullet \,\vec r_{23}} = \sqrt{(-0.500\,m)^{2}+(0.500\,m)^{2}}

r_{23} =\frac{\sqrt{2}}{2}\,m

The resultant force is found by Coulomb's law and principle of superposition:

\vec R = \vec F_{13}+\vec F_{23} (1)

Please notice that particles with charges of <em>same</em> sign attract each other and particles with charges of <em>opposite</em> sign repeal each other.

\vec R = \frac{k\cdot q_{1}\cdot q_{3}}{r_{13}^{2}}\cdot \vec u_{13}  +\frac{k\cdot q_{2}\cdot q_{3}}{r_{23}^{2}}\cdot \vec u_{23} (2)

Where:

  • k - Electrostatic constant, in newton-square meters per square Coulomb.
  • q_{1}, q_{2}, q_{3} - Electric charges, in Coulombs.
  • r_{13}, r_{23} - Distances between particles, in meters.
  • \vec u_{13}, \vec u_{23} - Unit vectors, no unit.

If we know that k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}, q_{1} = 2\times 10^{-6}\,C, q_{2} = 2\times 10^{-6}\,C, q_{3} = 2\times 10^{-6}\,C, r_{13} =\frac{\sqrt{2}}{2}\,m, r_{23} =\frac{\sqrt{2}}{2}\,m, \vec u_{13} = \left(-\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}  \right) and \vec u_{23} = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right), then the vector force on charge q_{3} is:

\vec R = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right) + \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right)

\vec R = 0.072\cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right) + 0.072\cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right)\,[N]

\vec R = 0.072\cdot \left(0, -\sqrt{2}\right)\,[N]

And the magnitude of the <em>electrical</em> force on charge q_{3} (R), in newtons, due to the others is found by Pythagorean theorem:

R = 0.102\,N

The magnitude of <em>electrical</em> force on charge q_{3} due to the others is 0.102 newtons. \blacksquare

To learn more on Coulomb's law, we kindly invite to check this verified question: brainly.com/question/506926

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  • For diagram refer the attachment.

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