(a) Find the total Coulomb force on a charge of \(2.00{\rm{ nC}}\) located at \(x = 4.00{\rm{ cm}}\) in Figure 18.52 (b), given that \(q = 1.00{\rm{ \mu C}}\). (b) Find the \({\rm{x}}\)-position at which the electric field is zero in Figure 18.52 (b).

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

The force of attraction or repulsion experienced by the charge particle when placed in an electric field is known as electrostatic force. The expression for the electrostatic force is,

\({\rm{F = }}\frac{{{\rm{KqQ}}}}{{{{\rm{r}}^{\rm{2}}}}}\)

Here, \({\rm{K}}\) is the electrostatic force constant, \({\rm{q}}\) and \({\rm{Q}}\) are two-point charges separated by the distance \({\rm{r}}\).

When the charge particle is placed on the system of charges, it experiences the electrostatic force due to all other particle in the system. The total force experienced by the charge is given as the vector of all the individual force acting on it.

The total Coulomb force acting on the charge at \(x = 4.00{\rm{ cm}}\) is represented as,

Total Coulomb force acting on the charge at \(x = 4.00{\rm{ cm}}\)

The total Coulomb force acting on the charge at \(x = 4.00{\rm{ cm}}\) is,

\(\begin{array}{c}F = - {F_{4,1}} - {F_{4,5}} - {F_{4,8}} + {F_{4,14}}\\ = - \frac{{K\left( {2q} \right)Q}}{{{{\left| {{r_4} - {r_1}} \right|}^2}}} - \frac{{K\left( q \right)Q}}{{{{\left| {{r_4} - {r_5}} \right|}^2}}} - \frac{{K\left( {3q} \right)Q}}{{{{\left| {{r_4} - {r_8}} \right|}^2}}} + \frac{{K\left( q \right)Q}}{{{{\left| {{r_4} - {r_{14}}} \right|}^2}}}\\ = KqQ\left[ { - \frac{2}{{{{\left| {{r_4} - {r_1}} \right|}^2}}} - \frac{1}{{{{\left| {{r_4} - {r_5}} \right|}^2}}} - \frac{3}{{{{\left| {{r_4} - {r_8}} \right|}^2}}} + \frac{1}{{{{\left| {{r_4} - {r_{14}}} \right|}^2}}}} \right]\end{array}\)

Here, \(K\) is the electrostatic force constant, \(q\) is the charge, \(Q\) is the test charge, \(\left| {{r_4} - {r_1}} \right|\) is the distance between the charges at \(x = 4.00{\rm{ cm}}\) and \(x = 1.00{\rm{ cm}}\), \(\left| {{r_4} - {r_5}} \right|\) is the distance between the charges at \(x = 4.00{\rm{ cm}}\) and \(x = 5.00{\rm{ cm}}\) \(\left| {{r_4} - {r_8}} \right|\) is the distance between the charges at \(x = 4.00{\rm{ cm}}\) and \(x = 8.00{\rm{ cm}}\), and \(\left| {{r_4} - {r_{14}}} \right|\) is the distance between the charges at \(x = 4.00{\rm{ cm}}\) and \(x = 14.00{\rm{ cm}}\).

Substitute \(9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}\) for \(K\), \[1{\rm{ \mu C}}\] for \(q\), \[2{\rm{ nC}}\] for \(Q\), \[1{\rm{ cm}}\] for \(r\), \[{\rm{4 cm}}\] for \({r_4}\), \[{\rm{5 cm}}\] for \({r_5}\), \[{\rm{8 cm}}\] for \({r_8}\), and \[14{\rm{ cm}}\] for \({r_{14}}\).

\[\begin{array}{c}F = \left( {9 \times {{10}^9}{\rm{ }}N \cdot {m^2}/{C^2}} \right) \times \left( {1{\rm{ }}\mu C} \right) \times \left( {2{\rm{ }}nC} \right) \times \left[ \begin{array}{l} - \frac{2}{{{{\left| {\left( {4{\rm{ }}cm} \right) - \left( {1{\rm{ }}cm} \right)} \right|}^2}}} - \frac{1}{{{{\left| {\left( {4{\rm{ }}cm} \right) - \left( {5{\rm{ }}cm} \right)} \right|}^2}}} - \\\frac{3}{{{{\left| {\left( {4{\rm{ }}cm} \right) - \left( {8{\rm{ }}cm} \right)} \right|}^2}}} + \frac{1}{{{{\left| {\left( {4{\rm{ }}cm} \right) - \left( {14{\rm{ }}cm} \right)} \right|}^2}}}\end{array} \right]\\ = \left( {9 \times {{10}^9}{\rm{ }}N \cdot {m^2}/{C^2}} \right) \times \left( {1 \times {{10}^{ - 6}}{\rm{ }}C} \right) \times \left( {2 \times {{10}^{ - 9}}{\rm{ }}C} \right) \times \left[ \begin{array}{l} - \frac{2}{{{{\left| {\left( {0.03{\rm{ }}m} \right)} \right|}^2}}} - \frac{1}{{{{\left| { - \left( {0.01{\rm{ }}cm} \right)} \right|}^2}}} - \\\frac{3}{{{{\left| { - \left( {0.04{\rm{ }}m} \right)} \right|}^2}}} + \frac{1}{{{{\left| { - \left( {0.1{\rm{ }}m} \right)} \right|}^2}}}\end{array} \right]\\ = - 0.252{\rm{ }}N\end{array}\]

Hence, the magnitude of total Coulomb force on the charge located at \(x = 4.00{\rm{ cm}}\) is \[0.252{\rm{ }}N\], and is directed towards left.

Let at \(x\) the electric field be zero.

The electric field at \(x\) is represented as,

Electric field at \(x\)

The electric field at \(x\) is,

\(\begin{array}{c}E = - {E_1} + {E_5} - {E_8} + {E_{14}}\\E = - \frac{{K\left( {2q} \right)}}{{{{\left( {x - {r_1}} \right)}^2}}} + \frac{{K\left( q \right)}}{{{{\left( {x - {r_5}} \right)}^2}}} - \frac{{K\left( {3q} \right)}}{{{{\left( {{r_8} - x} \right)}^2}}} + \frac{{K\left( q \right)}}{{{{\left( {{r_{14}} - x} \right)}^2}}}\\E = Kq\left[ { - \frac{2}{{{{\left( {x - {r_1}} \right)}^2}}} + \frac{1}{{{{\left( {x - {r_5}} \right)}^2}}} - \frac{3}{{{{\left( {{r_8} - x} \right)}^2}}} + \frac{1}{{{{\left( {{r_{14}} - x} \right)}^2}}}} \right]\\\frac{E}{{Kq}} = - \frac{2}{{{{\left( {x - {r_1}} \right)}^2}}} + \frac{1}{{{{\left( {x - {r_5}} \right)}^2}}} - \frac{3}{{{{\left( {{r_8} - x} \right)}^2}}} + \frac{1}{{{{\left( {{r_{14}} - x} \right)}^2}}}\end{array}\)

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

Substitute \(1.00{\rm{ cm}}\) for \({r_1}\), \(5.00{\rm{ cm}}\) for \({r_5}\), \(8.00{\rm{ cm}}\) for \({r_8}\), and \(14.00{\rm{ cm}}\) for \({r_{14}}\).

\(\frac{E}{{Kq}} = - \frac{2}{{{{\left[ {x - \left( {1.00{\rm{ }}cm} \right)} \right]}^2}}} + \frac{1}{{{{\left[ {x - \left( {5.00{\rm{ }}cm} \right)} \right]}^2}}} - \frac{3}{{{{\left[ {\left( {8.00{\rm{ }}cm} \right) - x} \right]}^2}}} + \frac{1}{{{{\left[ {\left( {14.00{\rm{ }}cm} \right) - x} \right]}^2}}}\)

Plotting this polynomial,

Plot for \(\frac{E}{{Kq}}\) vs x

From graph, it is clear that the electric field is zero at \(x = 6.073{\rm{ cm}}\) between \(x = 5.00{\rm{ cm}}\) and \(x = 8.00{\rm{ cm}}\).

Hence, the electric field is zero at \(x = 6.073{\rm{ cm}}\) between \(x = 5.00{\rm{ cm}}\) and \(x = 8.00{\rm{ cm}}\).