91影视

The force experienced by the two charges is due to the repulsion and attraction between them.

In this question, two electric forces act on the charges. One is the Coulomb force between the two charges and the second force is due to the electric field.The net electrostatic force is the sum of the two electric forces.

The given data can be listed below as:

• The distance between the positive and the negative charges is\(x = 0.47{\rm{ m}}\).
• The electric field produced by the charged plates is\(E = 53000{\rm{ N/C}}\).
• The value of the negative charge is\({Q_1} = - 6.7 \mu {\rm{C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) = - 6.7 \times {10^{ - 6}}{\rm{ C}}\).
• The value of the positive charge is\({Q_2} = 1.8{\rm{ }}\mu {\rm{C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) = 1.8 \times {10^{ - 6}}{\rm{ C}}\).
• The Coulomb鈥檚 law constant value is \(k = 9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}\).

According to Coulomb鈥檚 law, the force acting between the two charges can be expressed as:

\({F_1} = k\frac{{\left| {{Q_1}{Q_2}} \right|}}{{{{\left( x \right)}^2}}}\) 鈥� (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:

\({F_2} = \left| {{Q_1}} \right|E\) 鈥� (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.