Question

(a) Using the symmetry of the arrangement, determine the direction of the electric field at the center of the square in Figure 18.53, given that\({q_a} = {q_b} = - {\rm{1}}{\rm{.00 }}\mu {\rm{C}}\)and\({q_c} = {q_d} = + {\rm{1}}{\rm{.00 mC}}\). (b) Calculate the magnitude of the electric field at the location of\(q\), given that the square is\(5.00{\rm{ cm}}\)on a side.

Short answer

(a) The electric field at the center of the square will be straight up.

(b) The magnitude of the electric field at the center of the square is \({\rm{2}}{\rm{.03 \times 1}}{{\rm{0}}^{\rm{7}}}{\rm{ N/C}}\).

Step-by-step solution

Electric field

The space around the charge in which another charge experiences some electrostatic force is known as electric field. It is a vector quantity. The expression for the electric field is,

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

Here, \(K\) is the electrostatic force constant, \(q\) is the charge and \(r\) is the distance of the point of consideration from the charge \(q\).

If a number of electric field acts a given point, the net electric field will be equal to the vector sum of individual field.

(a) Direction of the electric field

Due to symmetry the electric field at the center of the square will be straight up, since \({q_a}\) and \({q_b}\) are negative and \({q_c}\) and \({q_d}\) are positive with same magnitude. The resultant of the electric field due to \({q_a}\) and \({q_b}\) will be straight up, and the resultant of the electric field due to \({q_c}\) and \({q_d}\) will be straight up.

Hence, the direction of electric field at the center of the square is straight up.

(b) Magnitude of the electric field at the center of square

The electric field at the center of the square is represented as,

Electric field at the center of the square

Here, \({E_a}\) is the electric field at the center of the square due to charge \({q_a}\), \({E_b}\) is the electric field at the center of the square due to charge \({q_b}\), \({E_c}\) is the electric field at the center of the square due to charge \({q_c}\), and \({E_d}\) is the electric field at the center of the square due to charge \({q_d}\).

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 \(\left( {a = {\rm{5}}{\rm{.00 cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}r = \frac{{5.00{\rm{ cm}}}}{{\sqrt 2 }}\\ \approx 3.54{\rm{ cm}}\end{array}\)

Electric field at the center of the square

The magnitude of the electric field at the center of the square due to charge \({q_a}\),

\({E_a} = \frac{{K{q_a}}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant \(\left( {K = {\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \({q_a}\) is the magnitude of the charge \(\left( {{q_a} = {\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right)\), and \(r\) is the distance of charge \({q_a}\) from the center of the square \(\left( {r = 3.54{\rm{ cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{E_a} = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{3}}{\rm{.54 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 \mu C}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 \mu C}}}}} \right)}}{{{{\left[ {\left( {{\rm{3}}{\rm{.54 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = {\rm{7}}{\rm{.18}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{ N/C}}\end{array}\)

The magnitude of the electric field at the center of the square due to charge \({q_b}\),

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

Here, \(K\) is the electrostatic force constant \(\left( {K = {\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \({q_b}\) is the magnitude of the charge \(\left( {{q_b} = 1.00{\rm{ }}\mu {\rm{C}}} \right)\), and \(r\) is the distance of charge \({q_b}\) from the center of the square \(\left( {r = 3.54{\rm{ cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{E_b} = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{3}}{\rm{.54 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 \mu C}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 \mu C}}}}} \right)}}{{{{\left[ {\left( {{\rm{3}}{\rm{.54 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = {\rm{7}}{\rm{.18}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{ N/C}}\end{array}\)

The magnitude of the electric field at the center of the square due to charge \({q_c}\),

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

Here, \(K\) is the electrostatic force constant \(\left( {K = {\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \({q_c}\) is the magnitude of the charge \(\left( {{q_c} = 1.00{\rm{ }}\mu {\rm{C}}} \right)\), and \(r\) is the distance of charge \({q_c}\) from the center of the square \(\left( {r = 3.54{\rm{ cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{E_c} = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{3}}{\rm{.54 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 \mu C}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 \mu C}}}}} \right)}}{{{{\left[ {\left( {{\rm{3}}{\rm{.54 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = {\rm{7}}{\rm{.18}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{ N/C}}\end{array}\)

The magnitude of the electric field at the center of the square due to charge \({q_d}\),

\({E_d} = \frac{{K{q_d}}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant \(\left( {K = {\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \({q_d}\) is the magnitude of the charge \(\left( {{q_d} = 1.00{\rm{ }}\mu {\rm{C}}} \right)\), and \(r\) is the distance of charge \({q_d}\) from the center of the square \(\left( {r = 3.54{\rm{ cm}}} \right)\).

Substituting all known values,

\(\begin{array}{c}{E_d} = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 }}\mu {\rm{C}}} \right)}}{{{{\left( {{\rm{3}}{\rm{.54 cm}}} \right)}^{\rm{2}}}}}\\ = \frac{{\left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( {{\rm{1}}{\rm{.00 \mu C}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 \mu C}}}}} \right)}}{{{{\left[ {\left( {{\rm{3}}{\rm{.54 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right]}^{\rm{2}}}}}\\ = {\rm{7}}{\rm{.18}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{ N/C}}\end{array}\)

Calculating the magnitude of net field

The field in the horizontal direction is,

\(\begin{array}{c}{E_x} = - {E_a}\sin \left( {45^\circ } \right) + {E_b}\sin \left( {45^\circ } \right) + {E_c}\sin \left( {45^\circ } \right) - {E_d}\sin \left( {45^\circ } \right)\\ = \left( { - {E_a} + {E_b} + {E_c} - {E_d}} \right) \times \sin \left( {45^\circ } \right)\end{array}\)

Substituting all known values,

\(\begin{array}{c}{E_x} = \left[ \begin{array}{l} - \left( {{\rm{7}}{\rm{.18}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{ N/C}}} \right) + \left( {{\rm{7}}{\rm{.18}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{ N/C}}} \right)\\ + \left( {{\rm{7}}{\rm{.18}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{ N/C}}} \right) - \left( {{\rm{7}}{\rm{.18}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{ N/C}}} \right)\end{array} \right] \times {\rm{sin}}\left( {{\rm{45^\circ }}} \right)\\ = 0\end{array}\)

The field in the vertical direction is,

\(\begin{array}{c}{E_y} = {E_a}\cos \left( {45^\circ } \right) + {E_b}\cos \left( {45^\circ } \right) + {E_c}\cos \left( {45^\circ } \right) + {E_d}\cos \left( {45^\circ } \right)\\ = \left( {{E_a} + {E_b} + {E_c} + {E_d}} \right) \times \cos \left( {45^\circ } \right)\end{array}\)

Substituting all known values,

\(\begin{array}{c}{E_y} = \left[ \begin{array}{l}\left( {{\rm{7}}{\rm{.18}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{ N/C}}} \right) + \left( {{\rm{7}}{\rm{.18}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{ N/C}}} \right) + \\\left( {{\rm{7}}{\rm{.18}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{ N}}/{\rm{C}}} \right) + \left( {{\rm{7}}{\rm{.18}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{ N/C}}} \right)\end{array} \right] \times {\rm{cos}}\left( {{\rm{45^\circ }}} \right)\\ = {\rm{2}}{\rm{.03}} \times {\rm{1}}{{\rm{0}}^{\rm{7}}}{\rm{ N}}/{\rm{C}}\end{array}\)

The magnitude of net electric field at the center of the square is,

\(E = \sqrt {E_x^2 + E_y^2} \)

Substituting all known values,

\(\begin{array}{c}E = \sqrt {{{\left( 0 \right)}^2} + {{\left( {2.03 \times {{10}^7}{\rm{ N}}/{\rm{C}}} \right)}^2}} \\ = 2.03 \times {10^7}{\rm{ N}}/{\rm{C}}\end{array}\)

Hence, the magnitude of electric field at the center of the square is \({\rm{2}}{\rm{.03}} \times {\rm{1}}{{\rm{0}}^{\rm{7}}}{\rm{ N}}/{\rm{C}}\).