Using the symmetry of the arrangement, determine the direction of the force on\(q\)in the figure below, given that\({q_a} = {q_b} = + 7.50{\rm{ }}\mu {\rm{C}}\)and\({q_c} = {q_d} = - 7.50{\rm{ }}\mu {\rm{C}}\). (b) Calculate the magnitude of the force on the charge\(q\), given that the square is\(10.0{\rm{ cm}}\)on a side and\(q = {\rm{2}}{\rm{.00 }}\mu {\rm{C}}\).

Electrostatic force is a vector quantity. When two charges are separated by some distance the electrostatic force between them is given as,

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

Here, \(K\)is the electrostatic force constant.

The net electrostatic force on the charge will be the vector sum of the individual charge.

Due to symmetry the net force on \(q\) will be straight down, since \({q_a}\) and \({q_b}\) are positive and \({q_c}\) and \({q_d}\) are negative with same magnitude. \({q_a}\) and \({q_b}\) will force the charge straight downward and \({q_c}\) and \({q_d}\) will pull the charge straight downward.

Hence, the direction of the force on \(q\) is straight downward.

The force on the charge \(q\) is represented as,

Force on the charge \(q\)

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

The distance of 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 \(\left( {a = 10.0{\rm{ }}cm} \right)\).

Substituting all known values,

\(\begin{array}{c}r = \frac{{10.0{\rm{ cm}}}}{{\sqrt 2 }}\\ = 7.07{\rm{ cm}}\end{array}\)

The magnitude of the repulsive force on charge \(q\) due to \({q_a}\) is,

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

Here, \(K\) is the electrostatic force constant \(\left( {K = {\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the magnitude of the charge at the center of the square \(\left( {q = 2.00{\rm{ }}\mu {\rm{C}}} \right)\), \({q_a}\) is the magnitude of the charge at the edge of the square \(\left( {{q_a} = 7.50{\rm{ }}\mu {\rm{C}}} \right)\), and \(r\) is the distance between \(q\) and \({q_a}\) \(\left( {r = 7.07{\rm{ cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{F_a} = \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 \mu C}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{7}}{\rm{.07 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)}}{{{{\left[ {\left( {{\rm{7}}{\rm{.07 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = {\rm{27 N}}\end{array}\)

The magnitude of the repulsive force on charge \(q\) due to \({q_b}\) is,

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

Here, \(K\) is the electrostatic force constant \(\left( {K = {\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the magnitude of the charge at the center of the square \(\left( {q = 2.00{\rm{ }}\mu {\rm{C}}} \right)\), \({q_b}\) is the magnitude of the charge at the edge of the square \(\left( {{q_b} = {\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)\), and \(r\) is the distance between \(q\) and \({q_b}\) \(\left( {r = 7.07{\rm{ cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{F_b} &= \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{7}}{\rm{.07 cm}}} \right)}^{\rm{2}}}}}\\ &= \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N \times }}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)}}{{{{\left[ {\left( {{\rm{7}}{\rm{.07 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\& = {\rm{27 N}}\end{array}\)

The magnitude of the attractive force on charge \(q\) due to \({q_c}\) is,

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

Here, \(K\) is the electrostatic force constant \(\left( {K = {\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the magnitude of the charge at the center of the square \(\left( {q = 2.00{\rm{ }}\mu {\rm{C}}} \right)\), \({q_c}\) is the magnitude of the charge at the edge of the square \(\left( {{q_c} = {\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)\), and \(r\) is the distance between \(q\) and \({q_c}\) \(\left( {r = 7.07{\rm{ cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{F_c} = \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{7}}{\rm{.07 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)}}{{{{\left[ {\left( {{\rm{7}}{\rm{.07 cm}}} \right){\rm{ \times }}\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\{\rm{ = 27 N}}\end{array}\)

The magnitude of the attractive force on charge \(q\) due to \({q_d}\) is,

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

Here, \(K\) is the electrostatic force constant \(\left( {K = {\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the magnitude of the charge at the center of the square \(\left( {q = 2.00{\rm{ }}\mu {\rm{C}}} \right)\), \({q_d}\) is the magnitude of the charge at the edge of the square \(\left( {{q_d} = {\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)\), and \(r\) is the distance between \(q\) and \({q_d}\) \(\left( {r = 7.07{\rm{ cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{F_d} = \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{7}}{\rm{.07 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{2}}{\rm{.00 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) \times \left( {{\rm{7}}{\rm{.50 }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)}}{{{{\left[ {\left( {{\rm{7}}{\rm{.07 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = {\rm{27 N}}\end{array}\)

The force on the horizonal direction 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,

\(\begin{array}{c}{F_x} = \left[ {\left( {{\rm{27 N}}} \right) - \left( {{\rm{27 N}}} \right) - \left( {{\rm{27 N}}} \right) + \left( {{\rm{27 N}}} \right)} \right] \times {\rm{sin}}\left( {{\rm{45^\circ }}} \right)\\ = 0\end{array}\)

The force on the vertical direction 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,

\(\begin{array}{c}{F_y} = \left[ {\left( {{\rm{27 N}}} \right){\rm{ + }}\left( {{\rm{27 N}}} \right){\rm{ + }}\left( {{\rm{27 N}}} \right){\rm{ + }}\left( {{\rm{27 N}}} \right)} \right] \times {\rm{cos}}\left( {{\rm{45^\circ }}} \right)\\ = {\rm{76}}{\rm{.37 N}}\end{array}\)

The magnitude of net force on charge \(q\) is,

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

Substituting all known values,

\(\begin{array}{c}F = \sqrt {{{\left( {\rm{0}} \right)}^{\rm{2}}}{\rm{ + }}{{\left( {{\rm{76}}{\rm{.37 N}}} \right)}^{\rm{2}}}} \\ = {\rm{76}}{\rm{.37 N}}\end{array}\)\(\)

Hence, the magnitude of the net force on the charge \(q\) is \(76.37{\rm{ N}}\).