91Ó°ÊÓ

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.

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 \({q_d}\).

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,

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

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,

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

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.