According to Coulomb’s law, the force acting between the two charges can be expressed as:
\({F_1} = k\frac{{\left| {{Q_1}{Q_2}} \right|}}{{{{\left( x \right)}^2}}}\) … (i)
Here, the electric force acts in the right direction along the positive x-axis. This force is exerted by charge\({Q_2}\)on\({Q_1}\). So,\({F_1}\)is positive.
The magnitude of the electric force acting on charge\({Q_1}\)can be expressed as:
\({F_2} = \left| {{Q_1}} \right|E\) … (ii)
Here, the force acts on\({Q_1}\)in the left direction along the negative x-axis. This force is due to the parallel plates. So,\({F_2}\)is negative.
Add equations (i) and (ii). The net electrostatic force on charge\({Q_1}\)can be expressed as:
\(\begin{aligned}{c}F &= {F_1} + \left( { - {F_2}} \right)\\ &= {F_1} - {F_2}\\ &= k\frac{{\left| {{Q_1}{Q_2}} \right|}}{{{{\left( x \right)}^2}}} - \left| {{Q_1}} \right|E\end{aligned}\)
Substitute the values in the above equation.
\(\begin{aligned}{c}F &= 9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2} \times \frac{{\left| {\left( { - 6.7 \times {{10}^{ - 6}}{\rm{ C}}} \right)\left( {1.8 \times {{10}^{ - 6}}{\rm{ C}}} \right)} \right|}}{{{{\left( {0.47{\rm{ m}}} \right)}^2}}} - \left| {\left( { - 6.7 \times {{10}^{ - 6}}{\rm{ C}}} \right)} \right| \times 53000{\rm{ N/C}}\\ &= 9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2} \times \frac{{\left( {6.7 \times {{10}^{ - 6}}{\rm{ C}}} \right)\left( {1.8 \times {{10}^{ - 6}}{\rm{ C}}} \right)}}{{{{\left( {0.47{\rm{ m}}} \right)}^2}}} - \left( {6.7 \times {{10}^{ - 6}}{\rm{ C}}} \right) \times 53000{\rm{ N/C}}\\ &= 0.491{\rm{ N}} - 0.355{\rm{ N}}\\ &= 0.136{\rm{ N}}\end{aligned}\)
The net force is positive. It means that the net force acting on the charge is in the right direction along the positive x-axis.
Thus, the magnitude of the net electrostatic force on \({Q_1}\) is \(0.136{\rm{ N}}\) and it is directed toward the right.
