Question

(a) Find the total electric field at\(x = 1.00{\rm{ cm}}\)in Figure 18.52(b) given that\(q = {\rm{5}}{\rm{.00 nC}}\). (b) Find the total electric field at\(x = {\rm{11}}{\rm{.00 cm}}\)in Figure 18.52(b). (c) If the charges are allowed to move and eventually be brought to rest by friction, what will the final charge configuration be? (That is, will there be a single charge, double charge, etc., and what will its value(s) be?)

Figure 18.52 (a) Point charges located at \({\bf{3}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{11}}.{\bf{0}}{\rm{ }}{\bf{cm}}\) along the x-axis. (b) Point charges located at \({\bf{1}}.{\bf{00}},{\rm{ }}{\bf{5}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{14}}.{\bf{0}}{\rm{ }}{\bf{cm}}\) along the x-axis

Short answer

(a) The total electric field at \(x = 1.00{\rm{ cm}}\) is \( - \infty \).

(b) The total electric field at \(x = 11.00{\rm{ cm}}\) is \(2.035 \times {10^5}{\rm{ N}}/{\rm{C}}\).

(c) The final charge of the configuration is \( + q\).

Step-by-step solution

Total electric field

Electric field is a vector quantity given as,

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

Here,\(K\)is the electrostatic force constant\(\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)\).

The total electric field created by system of multiple charges is the vector sum of the individual field created by each charge.

(a) Total electric field at x=1.00 cm

The total electric field created at \(x = 1.00{\rm{ cm}}\) is,

E = Kq/r²

Here, \(K\) is the electrostatic force constant, \(q\) is the charge, \(\left| {{r_1} - {r_1}} \right|\) is the distance of the charge at \(x = 1.00{\rm{ cm}}\) and \(x = 1.00{\rm{ cm}}\), \(\left| {{r_1} - {r_5}} \right|\) is the distance of the charge at \({\mathop{\rm x}\nolimits} = 5.00 cm\) and \(x = 1.00{\rm{ cm}}\), \(\left| {{r_1} - {r_8}} \right|\) is the distance of the charge at \(x = 8.00{\rm{ cm}}\) and \(x = 1.00{\rm{ cm}}\), and \(\left| {{r_1} - {r_{14}}} \right|\) is the distance of the charge at \(x = 14.00{\rm{ cm}}\) and .

Substitute for , \(5.00{\rm{ nC}}\) for \(q\), \(1.00{\rm{ cm}}\) for \({r_1}\), \({\rm{5}}{\rm{.00 cm}}\) for \({r_5}\), \({\rm{8}}{\rm{.00 cm}}\) for \({r_8}\), and \(14.00{\rm{ cm}}\) for \({r_{14}}\).

\(\begin{array}{c}E = \left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right) \times \left[ \begin{array}{l}\frac{{\left( { - 2 \times 5{\rm{ nC}}} \right)}}{{{{\left| {\left( {1{\rm{ cm}}} \right) - \left( {1{\rm{ cm}}} \right)} \right|}^2}}} + \frac{{\left( {5{\rm{ nC}}} \right)}}{{{{\left| {\left( {1{\rm{ cm}}} \right) - \left( {5{\rm{ cm}}} \right)} \right|}^2}}}\\ + \frac{{\left( {3 \times 5{\rm{ nC}}} \right)}}{{{{\left| {\left( {1{\rm{ cm}}} \right) - \left( {8{\rm{ cm}}} \right)} \right|}^2}}} + \frac{{\left( { - 1 \times 5{\rm{ nC}}} \right)}}{{{{\left| {\left( {1{\rm{ cm}}} \right) - \left( {14{\rm{ cm}}} \right)} \right|}^2}}}\end{array} \right]\\ = - \infty \end{array}\)

Hence, the total electric field at \(x = 1.00{\rm{ cm}}\) is \( - \infty \).

(b) Total electric field at x=11.00 cm

The total electric field created at \(x = 11.00{\rm{ cm}}\) is represented as,

Electric field at \(x = 11.00{\rm{ cm}}\)

The total electric field created at \(x = 11.00{\rm{ cm}}\) is,

\(\begin{array}{c}E = - {E_1} + {E_5} + {E_8} + {E_{14}}\\ = - \frac{{K\left( {2q} \right)}}{{{{\left| {{r_{11}} - {r_1}} \right|}^2}}} + \frac{{K\left( q \right)}}{{{{\left| {{r_{11}} - {r_5}} \right|}^2}}} + \frac{{K\left( {3q} \right)}}{{{{\left| {{r_{11}} - {r_8}} \right|}^2}}} + \frac{{K\left( q \right)}}{{{{\left| {{r_{11}} - {r_{14}}} \right|}^2}}}\\ = K\left( { - \frac{{\left( {2q} \right)}}{{{{\left| {{r_{11}} - {r_1}} \right|}^2}}} + \frac{{\left( q \right)}}{{{{\left| {{r_{11}} - {r_5}} \right|}^2}}} + \frac{{\left( {3q} \right)}}{{{{\left| {{r_{11}} - {r_8}} \right|}^2}}} + \frac{{\left( q \right)}}{{{{\left| {{r_{11}} - {r_{14}}} \right|}^2}}}} \right)\end{array}\)

Here, \(\frac{1}{2}\) is the electrostatic force constant, \(q\) is the charge, \(\left| {{r_{11}} - {r_1}} \right|\) is the distance of the charge at \(x = 11.00{\rm{ cm}}\) and \(x = 1.00{\rm{ cm}}\), \(\left| {{r_{11}} - {r_5}} \right|\) is the distance of the charge at \(x = 5.00{\rm{ cm}}\) and \(x = 11.00{\rm{ cm}}\), \(\left| {{r_{11}} - {r_8}} \right|\) is the distance of the charge at \(x = 8.00{\rm{ cm}}\) and \(x = 11.00{\rm{ cm}}\), and \(\left| {{r_{11}} - {r_{14}}} \right|\) is the distance of the charge at \(x = 14.00{\rm{ cm}}\) and \(x = 11.00{\rm{ cm}}\).

Substitute \(9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\) for \(K\), \(5.00{\rm{ nC}}\) for \(q\), \(1.00{\rm{ cm}}\) for \({r_1}\), \(5.00{\rm{ cm}}\) for \({r_5}\), \({\rm{8}}{\rm{.00 cm}}\) for \({r_8}\), \(11.00{\rm{ cm}}\) for \({r_{11}}\) and \(14.00{\rm{ cm}}\) for \({r_{14}}\).

\(\begin{array}{c}E = \left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right) \times \left[ \begin{array}{l} - \frac{{\left( {2 \times 5{\rm{ nC}}} \right)}}{{{{\left| {\left( {11{\rm{ cm}}} \right) - \left( {1{\rm{ cm}}} \right)} \right|}^2}}} + \frac{{\left( {5{\rm{ nC}}} \right)}}{{{{\left| {\left( {11{\rm{ cm}}} \right) - \left( {5{\rm{ cm}}} \right)} \right|}^2}}}\\ + \frac{{\left( {3 \times 5{\rm{ nC}}} \right)}}{{{{\left| {\left( {11{\rm{ cm}}} \right) - \left( {8{\rm{ cm}}} \right)} \right|}^2}}} + \frac{{\left( {5{\rm{ nC}}} \right)}}{{{{\left| {\left( {11{\rm{ cm}}} \right) - \left( {14{\rm{ cm}}} \right)} \right|}^2}}}\end{array} \right]\\ = \left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right) \times \left[ \begin{array}{l} - \frac{{\left( {10{\rm{ nC}}} \right) \times \left( {\frac{{{{10}^{ - 9}}{\rm{ C}}}}{{1{\rm{ nC}}}}} \right)}}{{{{\left| {\left( {10{\rm{ cm}}} \right) \times \left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right|}^2}}} + \frac{{\left( {5{\rm{ nC}}} \right) \times \left( {\frac{{{{10}^{ - 9}}{\rm{ C}}}}{{1{\rm{ nC}}}}} \right)}}{{{{\left| {\left( {6{\rm{ cm}}} \right) \times \left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right|}^2}}}\\ + \frac{{\left( {15{\rm{ nC}}} \right) \times \left( {\frac{{{{10}^{ - 9}}{\rm{ C}}}}{{1{\rm{ nC}}}}} \right)}}{{{{\left| {\left( {3{\rm{ cm}}} \right) \times \left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right|}^2}}} + \frac{{\left( {5{\rm{ nC}}} \right) \times \left( {\frac{{{{10}^{ - 9}}{\rm{ C}}}}{{1{\rm{ nC}}}}} \right)}}{{{{\left| { - \left( {3{\rm{ cm}}} \right) \times \left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right|}^2}}}\end{array} \right]\\ = 2.035 \times {10^5}{\rm{ N}}/{\rm{C}}\end{array}\)

Hence, the total electric field at \(x = 11.00{\rm{ cm}}\) is \(2.035 \times {10^5}{\rm{ N}}/{\rm{C}}\).

(c) Net charge

According to the additive property of the charge, the net charge on the system is,

\(\begin{array}{c}{q_{net}} = - 2q + q + 3q - q\\ = + q\end{array}\)

Hence, the final charge of the configuration is \( + q\).