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Find the total Coulomb force on the charge \(q\) in Figure 18.53, given that \(q = {\rm{1}}{\rm{.00 }}\mu {\rm{C}}\), \({q_a} = {\rm{2}}{\rm{.00 }}\mu {\rm{C}}\), \({q_b} = - {\rm{3}}{\rm{.00 }}\mu {\rm{C}}\), \({q_c} = - {\rm{4}}{\rm{.00 }}\mu {\rm{C}}\) , and \({q_d} = + {\rm{1}}{\rm{.00 }}\mu {\rm{C}}\). The square is \({\rm{50}}{\rm{.0 cm}}\) on a side.

Short Answer

Expert verified

The total Coulomb force on the charge \(q\) is \({\rm{0}}{\rm{.102 N}}\) and is directed downwards.

Step by step solution

01

Coulomb force

When two charged particles are separated by some distance, they try to attract or repel each other by a force known as Coulomb force or electrostatic force. It is a vector quantity.

The expression for the magnitude of the Coulomb force is,

\(F = \frac{{KqQ}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant, \(q\) and \(Q\) are the charges separated by a distance \(r\).

For a system of charge, the total Coulomb force can be calculated by taking the vector sum of all the Coulomb force acting on a particle.

02

Force acting on the charge located at the center of the square

The force acting on the charge \(q\)located at the center of the square is represented as,

Force acting on the charge \(q\)

Here, \({F_a}\) is the repulsive force due to positive charge \({q_a}\), \({F_b}\) is the attractive force due to negative charge \({q_b}\), \({F_c}\) is the attractive force due to negative charge \({q_c}\), and \({F_d}\) is the repulsive force due to positive charge \({q_d}\).

03

Distance of charges from the center of the square

The distance of test charge \(q\) from the charges \({q_a}\), \({q_b}\), \({q_c}\) and \({q_d}\) is,

\(r = \frac{a}{{\sqrt 2 }}\)

Here, \(a\) is the side of the square \(\left( {a = {\rm{50}}{\rm{.0 cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}r = \frac{{{\rm{50}}{\rm{.0 cm}}}}{{\sqrt {\rm{2}} }}\\ = {\rm{35}}{\rm{.36 cm}}\end{array}\)

04

Calculating individual forces

The repulsive force due to positive charge \({q_a}\) is,

\({F_a} = \frac{{Kq{q_a}}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant \(\left( {K = {\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the magnitude of the charge at the center of the square \(\left( {q = {\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right)\), \({q_a}\) is the magnitude of the charge \(\left( {{q_a} = {\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right)\), and \(r\) is the distance between the charges \(q\) and \({q_a}\) \(\left( {r = {\rm{35}}{\rm{.36 cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{F_a} = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{35}}{\rm{.36 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 }}\mu {\rm{C}}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 }}\mu {\rm{C}}}}} \right)}}{{{{\left[ {\left( {{\rm{35}}{\rm{.36 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = {\rm{0}}{\rm{.144 N}}\end{array}\)

The attractive force due to negative charge \({q_b}\) is,

\({F_b} = \frac{{Kq{q_b}}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant \(\left( {K = {\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the magnitude of the charge at the center of the square \(\left( {q = {\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right)\), \({q_b}\) is the magnitude of the charge \(\left( {{q_b} = {\rm{3}}{\rm{.00 }}\mu {\rm{C}}} \right)\), and \(r\) is the distance between the charges \(q\) and \({q_b}\) \(\left( {r = {\rm{35}}{\rm{.36 cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{F_b} = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {{\rm{3}}{\rm{.00 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{35}}{\rm{.36 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 }}\mu {\rm{C}}}}} \right){\rm{ \times }}\left( {{\rm{3}}{\rm{.00 }}\mu {\rm{C}}} \right){\rm{ \times }}\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 }}\mu {\rm{C}}}}} \right)}}{{{{\left[ {\left( {{\rm{35}}{\rm{.36 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = {\rm{0}}{\rm{.216 N}}\end{array}\)

The attractive force due to negative charge \({q_c}\) is,

\({F_c} = \frac{{Kq{q_c}}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant \(\left( {K = {\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the magnitude of the charge at the center of the square \(\left( {q = {\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right)\), \({q_c}\) is the magnitude of the charge \(\left( {{q_c} = 4.00{\rm{ }}\mu {\rm{C}}} \right)\), and \(r\) is the distance between the charges \(q\) and \({q_c}\) \(\left( {r = {\rm{35}}{\rm{.36 cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{F_c} = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {{\rm{4}}{\rm{.00 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{35}}{\rm{.36 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 }}\mu {\rm{C}}}}} \right) \times \left( {{\rm{4}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 }}\mu {\rm{C}}}}} \right)}}{{{{\left[ {\left( {{\rm{35}}{\rm{.36 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = 0.288{\rm{ N}}\end{array}\)

The repulsive force due to positive charge \({q_d}\) is,

\({F_d} = \frac{{Kq{q_d}}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant \(\left( {K = {\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the magnitude of the charge at the center of the square \(\left( {q = {\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right)\), \({q_d}\) is the magnitude of the charge \(\left( {{q_d} = 1.00{\rm{ }}\mu {\rm{C}}} \right)\), and \(r\) is the distance between the charges \(q\) and \({q_d}\) \(\left( {r = {\rm{35}}{\rm{.36 cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{F_d} = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {1.00{\rm{ }}\mu {\rm{C}}} \right) \times \left( {1.00{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {35.36{\rm{ cm}}} \right)}^2}}}\\ = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {1.00{\rm{ }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) \times \left( {1.00{\rm{ }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)}}{{{{\left[ {\left( {35.36{\rm{ cm}}} \right) \times \left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right]}^2}}}\\ = 0.072{\rm{ N}}\end{array}\)

05

Calculating total Coulomb force

The horizontal component of the force is,

\(\begin{array}{c}{F_x} = {F_a}\sin \left( {45^\circ } \right) + {F_b}\sin \left( {45^\circ } \right) - {F_c}\sin \left( {45^\circ } \right) - {F_d}\sin \left( {45^\circ } \right)\\ = \left( {{F_a} + {F_b} - {F_c} - {F_d}} \right) \times \sin \left( {45^\circ } \right)\end{array}\)

Substituting all known values,

\(\begin{array}{c}{F_x} = \left[ {\left( {{\rm{0}}{\rm{.144 N}}} \right) + \left( {{\rm{0}}{\rm{.216 N}}} \right) - \left( {{\rm{0}}{\rm{.288 N}}} \right) - \left( {{\rm{0}}{\rm{.072 N}}} \right)} \right] \times {\rm{sin}}\left( {{\rm{45^\circ }}} \right)\\ = 0\end{array}\)

The vertical component of the force is,

\(\begin{array}{c}{F_y} = - {F_a}\cos \left( {45^\circ } \right) + {F_b}\cos \left( {45^\circ } \right) - {F_c}\cos \left( {45^\circ } \right) + {F_d}\cos \left( {45^\circ } \right)\\ = \left( { - {F_a} + {F_b} - {F_c} + {F_d}} \right) \times \cos \left( {45^\circ } \right)\end{array}\)

Substituting all known values,

\(\begin{array}{c}{F_y} = \left[ { - \left( {{\rm{0}}{\rm{.144 N}}} \right) + \left( {{\rm{0}}{\rm{.216 N}}} \right) - \left( {{\rm{0}}{\rm{.288 N}}} \right) + \left( {{\rm{0}}{\rm{.072 N}}} \right)} \right] \times {\rm{cos}}\left( {{\rm{45^\circ }}} \right)\\ = - {\rm{0}}{\rm{.102 N}}\end{array}\)

The direction of the resultant force is,

\(\theta = {\tan ^{ - 1}}\left( {\frac{{{F_y}}}{{{F_x}}}} \right)\)

Substituting all known values,

\(\begin{array}{c}\theta = {\tan ^{ - 1}}\left( {\frac{{ - 0.102{\rm{ N}}}}{0}} \right)\\ = 270^\circ \end{array}\)

The magnitude of the resultant force is,

\(F = \sqrt {F_x^2 + F_y^2} \)

Substituting all known values,

\(\begin{array}{c}F = \sqrt {{{\left( 0 \right)}^2} + {{\left( { - 0.102{\rm{ N}}} \right)}^2}} \\ = 0.102{\rm{ N}}\end{array}\)

Hence, total Coulomb force on the charge \(q\) is \({\rm{0}}{\rm{.102 N}}\) and is directed downwards.

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