Question

(II) Particles of charge +65, +48, and -95 uC are placed in a line (Fig. 16–52). The center one is 0.35 m from each of the others. Calculate the net force on each charge due to the other two.

Short answer

The net force on each charge is\( - 115.6\;{\rm{N}}\),\(563.4\;{\rm{N}}\) and \( - 447.8\;{\rm{N}}\), respectively.

Step-by-step solution

Understanding the electric force

A point charge exerts an electric force on another point charge in the vicinity. It could be attractive as well as repulsive depending upon the nature of the two charges.

The expression for the electric force is given as:

\(F = k\frac{{{Q_1}{Q_2}}}{{{r^2}}}\) … (i)

Here, k is the Coulomb’s constant,\({Q_1},\;{Q_2}\)are the charges and r is the separation between the charges.

Given Data

The charge of particle 1 is \({q_1} = 65\;{\rm{\mu C}}\).

The charge of particle 2 is \({q_2} = 48\;{\rm{\mu C}}\).

The charge of particle 3 is \({q_3} = 95\;{\rm{\mu C}}\).

The separation distance between 1 and 2 is, \(r = 0.35\;{\rm{m}}\).

The total distance between 1 and 3 is, \({r_0} = 0.7\;{\rm{m}}\).

Determination of the electric force on charge 1

From equation (i), the expression to find the force on charge one is given as:

\({F_1} = \frac{{k{q_1}{q_2}}}{{{r^2}}} + \frac{{k{q_1}{q_3}}}{{{r_0}^2}}\)

Substitute the values in the above expression.

\(\begin{aligned}{l}{F_1} = - \left( {8.988 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\left( {\frac{{\left( {65 \times {{10}^{ - 6}}\;{\rm{C}}} \right)\left( {48 \times {{10}^{ - 6}}\;{\rm{C}}} \right)}}{{{{\left( {0.35\;{\rm{m}}} \right)}^2}}} + \frac{{\left( {65 \times {{10}^{ - 6}}\;{\rm{C}}} \right)\left( {95 \times {{10}^{ - 6}}\;{\rm{C}}} \right)}}{{{{\left( {0.7\;{\rm{m}}} \right)}^2}}}} \right)\\{F_1} = - 115.6\;{\rm{N}}\end{aligned}\)

Thus, the net force on charge 1 is \( - 115.6\;{\rm{N}}\) towards the left.

Determination of the electric force on charge 2

From equation (i), the expression to find the force on charge two is given as:

\({F_2} = \frac{{k{q_1}{q_2}}}{{{r^2}}} + \frac{{k{q_2}{q_3}}}{{{r^2}}}\)

Substitute the values in the above expression.

\(\begin{aligned}{l}{F_2} = \left( {8.988 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\left( {\frac{{\left( {65 \times {{10}^{ - 6}}\;{\rm{C}}} \right)\left( {48 \times {{10}^{ - 6}}\;{\rm{C}}} \right)}}{{{{\left( {0.35\;{\rm{m}}} \right)}^2}}} + \frac{{\left( {48 \times {{10}^{ - 6}}\;{\rm{C}}} \right)\left( {95 \times {{10}^{ - 6}}\;{\rm{C}}} \right)}}{{{{\left( {0.35\;{\rm{m}}} \right)}^2}}}} \right)\\{F_2} = 563.4\;{\rm{N}}\end{aligned}\)

Thus, the net force on charge 2 is \(563.4\;{\rm{N}}\) towards the right.

Determination of the electric force on charge 3

From equation (i), the expression to find the force on charge three is given as:

\({F_3} = \frac{{k{q_1}{q_3}}}{{{r_0}^2}} + \frac{{k{q_2}{q_3}}}{{{r^2}}}\)

Substitute the values in the above expression.

\(\begin{aligned}{l}{F_3} = - \left( {8.988 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\left( {\frac{{\left( {65 \times {{10}^{ - 6}}\;{\rm{C}}} \right)\left( {95 \times {{10}^{ - 6}}\;{\rm{C}}} \right)}}{{{{\left( {0.35\;{\rm{m}}} \right)}^2}}} + \frac{{\left( {48 \times {{10}^{ - 6}}\;{\rm{C}}} \right)\left( {95 \times {{10}^{ - 6}}\;{\rm{C}}} \right)}}{{{{\left( {0.35\;{\rm{m}}} \right)}^2}}}} \right)\\{F_3} = - 447.8\;{\rm{N}}\end{aligned}\)

Thus, the net force on charge 3 is \( - 447.8\;{\rm{N}}\) towards the left.