Question

Find the electric field at the location of \({q_a}\) in Figure 18.53 given that \({q_b} = {q_c} = {q_d} = + 2.00{\rm{ nC}}\), \(q = - 1.00{\rm{ nC}}\), and the square is \({\rm{20}}{\rm{.0 cm}}\) on a side.

Short answer

The electric field at the location of \({q_a}\) is \(411.4{\rm{ N/C}}\).

Step-by-step solution

Electric field

The electric field at point at a distance \(r\) from the charge \(q\) is,

\(E = \frac{{Kq}}{{{r^2}}}\)

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

For a system of charges, the electric field at point can be determined by the vector sum of all the field acting at that point.

Electric field at the location

The electric field at the location of \({q_a}\) is represented as,

Electric field at the location of \({q_a}\)

Here, \(E\) is the net electric field, \({E_b}\) is the electric field due to charge \({q_b}\), \({E_c}\) is the electric field due to charge \({q_c}\), \({E_{bc}}\) is the resultant electric field of \({E_b}\) and \({E_c}\), \({E_q}\) is the electric field due to charge \(q\), and \({E_d}\) is the electric field due to charge \({q_d}\).

Calculating electric fields due to individual charge

The electric field due to charge \({q_b}\),

\({E_b} = \frac{{K{q_b}}}{{{a^2}}}\)

Here, \(K\) is the electrostatic force constant \[\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\], \({q_b}\) is the magnitude of charge \(\left( {{q_b} = 2.00{\rm{ nC}}} \right)\), and a is the side of square \(\left( {a = 20.0{\rm{ cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{E_b} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {2.00{\rm{ }}nC} \right)}}{{{{\left( {20.0{\rm{ }}cm} \right)}^2}}}\\ = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {2.00{\rm{ }}nC} \right) \times \left( {\frac{{{{10}^{ - 9}}{\rm{ }}C}}{{1{\rm{ }}nC}}} \right)}}{{{{\left[ {\left( {20.0{\rm{ }}cm} \right) \times \left( {\frac{{{{10}^{ - 2}}{\rm{ }}m}}{{1{\rm{ }}cm}}} \right)} \right]}^2}}}\\ = 450{\rm{ N/C}}\end{array}\)

The electric field due to charge \({q_c}\),

\({E_c} = \frac{{K{q_c}}}{{{a^2}}}\)

Here, \(K\) is the electrostatic force constant \[\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\], \({q_c}\) is the magnitude of charge \(\left( {{q_c} = 2.00{\rm{ nC}}} \right)\), and a is the side of square \(\left( {a = 20.0{\rm{ cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{E_c} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {2.00{\rm{ }}nC} \right)}}{{{{\left( {20.0{\rm{ }}cm} \right)}^2}}}\\ = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {2.00{\rm{ }}nC} \right) \times \left( {\frac{{{{10}^{ - 9}}{\rm{ }}C}}{{1{\rm{ }}nC}}} \right)}}{{{{\left[ {\left( {20.0{\rm{ }}cm} \right) \times \left( {\frac{{{{10}^{ - 2}}{\rm{ }}m}}{{1{\rm{ }}cm}}} \right)} \right]}^2}}}\\ = 450{\rm{ N/C}}\end{array}\)

The electric field due to charge \(q\),

\({E_q} = \frac{{Kq}}{{{{\left( {\frac{a}{{\sqrt 2 }}} \right)}^2}}}\)

Here, \(K\) is the electrostatic force constant \[\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\], \({q_c}\) is the magnitude of charge \(\left( {q = 1.00{\rm{ nC}}} \right)\), and a is the side of square \(\left( {a = 20.0{\rm{ cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{E_q} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {1.00{\rm{ }}nC} \right)}}{{{{\left( {\frac{{20.0{\rm{ }}cm}}{{\sqrt 2 }}} \right)}^2}}}\\ = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {1.00{\rm{ }}nC} \right) \times \left( {\frac{{{{10}^{ - 9}}{\rm{ }}C}}{{1{\rm{ }}nC}}} \right)}}{{{{\left[ {{{\left( {\frac{{20.0{\rm{ }}cm}}{{\sqrt 2 }}} \right)}^2} \times \left( {\frac{{{{10}^{ - 2}}{\rm{ }}m}}{{1{\rm{ }}cm}}} \right)} \right]}^2}}}\\ = 450{\rm{ N/C}}\end{array}\)

The electric field due to charge \({q_d}\),

\({E_d} = \frac{{K{q_d}}}{{{{\left( {\sqrt 2 a} \right)}^2}}}\)

Here, \(K\) is the electrostatic force constant \[\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\], \({q_d}\) is the magnitude of charge \(\left( {{q_d} = 2.00{\rm{ nC}}} \right)\), and a is the side of square \(\left( {a = 20.0{\rm{ cm}}} \right)\).

Substituting all known values,

\[\begin{array}{c}{E_d} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {2.00{\rm{ }}nC} \right)}}{{{{\left( {\sqrt 2 \times 20.0{\rm{ }}cm} \right)}^2}}}\\ = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {2.00{\rm{ }}nC} \right) \times \left( {\frac{{{{10}^{ - 9}}{\rm{ }}C}}{{1{\rm{ }}nC}}} \right)}}{{{{\left[ {\sqrt 2 \times \left( {20.0{\rm{ }}cm} \right) \times \left( {\frac{{{{10}^{ - 2}}{\rm{ }}m}}{{1{\rm{ }}cm}}} \right)} \right]}^2}}}\\ = 225{\rm{ N/C}}\end{array}\]

Calculating the resultant electric field

The resultant of electric fields \[{E_b}\] and \[{E_c}\] is,

\[{E_{bc}} = \sqrt {E_b^2 + E_c^2 + 2{E_b}{E_c}\cos \theta } \]

Here, \[\theta \] is the angle between the electric fields \[{E_b}\] and \[{E_c}\] \(\left( {\theta = {{90}^ \circ }} \right)\).

Substituting all known values,

\(\begin{array}{c}{E_{bc}} = \sqrt {{{\left( {450{\rm{ }}N} \right)}^2} + {{\left( {450{\rm{ }}N} \right)}^2} + 2 \times \left( {450{\rm{ }}N} \right) \times \left( {450{\rm{ }}N} \right) \times \cos \left( {90^\circ } \right)} \\ = 636.4{\rm{ N/C}}\end{array}\)

The net electric field at the location of \({q_a}\) is,

\(E = {E_{bc}} + {E_d} - {E_q}\)

Substituting all known values,

\(\begin{array}{c}E = \left( {636.4{\rm{ N/C}}} \right) + \left( {225{\rm{ N/C}}} \right) - \left( {450{\rm{ N/C}}} \right)\\ = 411.4{\rm{ N/C}}\end{array}\)

Hence, the electric field at the location of \({q_a}\) is \(411.4{\rm{ N/C}}\).