Question

(II) Determine the magnitude and direction of the electric field at a point midway between a \( - {\bf{8}}{\bf{.0}}{\rm{ }}\mu {\bf{C}}\) and a\( + 5.8{\rm{ }}\mu {\bf{C}}\)charge 6.0 cm apart. Assume no other charges are nearby.

Short answer

The magnitude of the electric field at a point midway between both charges is \(1.38 \times {10^8}{\rm{ N/C}}\). The direction of an electric field is toward the negative charge

Step-by-step solution

Understanding the acceleration of an electron

When the electric force acts on an electron, the electron starts accelerating.The acceleration of an electron depends on the charge, electric field, and mass of an electron.The unit of acceleration of an electron is a meter per second square.

Identification of given data

The given data is listed below:

• The distance between the positive and negative charges is\(d = 6{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) = 0.06{\rm{ m}}\).
• The value of the positive charge is\({Q_1} = 5.8{\rm{ }}\mu {\rm{C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) = 5.8 \times {10^{ - 6}}{\rm{ C}}\).
• The value of the positive charge is\({Q_2} = - 8{\rm{ }}\mu {\rm{C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) = - 8 \times {10^{ - 6}}{\rm{ C}}\).
• The Coulomb’s law constant value is \(k = 9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}\).

Representation of the charges with the electric fields

Here, the distance of the positive and negative charges from the center of an electric field is half of the total distance, i.e., (d/2). The charge of\({Q_1}\)is positive and\({Q_2}\)is negative.

Here, \(\overrightarrow {{{\bf{E}}_{\bf{1}}}} \) and \(\overrightarrow {{{\bf{E}}_{\bf{2}}}} \)are the electric field vectors of charge \({Q_1}\) and \({Q_2}\), respectively.

Determination of the magnitude and the direction of an electric field at the midway between two charges

The magnitude of an electric field of charge\({Q_1}\)can be expressed as follows:

\({E_1} = k\frac{{\left| {{Q_1}} \right|}}{{{{\left( {\frac{d}{2}} \right)}^2}}}\) … (i)

The magnitude of an electric field of charge\({Q_2}\)can be expressed as follows:

\({E_2} = k\frac{{\left| {{Q_2}} \right|}}{{{{\left( {\frac{d}{2}} \right)}^2}}}\) … (ii)

Add equations (i) and (ii). The electric field at the mid of both charges can be expressed as follows:

\(\begin{aligned}{c}E = \left| {{E_1}} \right| + \left| {{E_2}} \right|\\ = k\frac{{\left| {{Q_1}} \right|}}{{{{\left( {\frac{d}{2}} \right)}^2}}} + k\frac{{\left| {{Q_2}} \right|}}{{{{\left( {\frac{d}{2}} \right)}^2}}}\\ = 4k\frac{{\left| {{Q_1}} \right|}}{{{d^2}}} + 4k\frac{{\left| {{Q_2}} \right|}}{{{d^2}}}\\ = = \frac{{4k}}{{{d^2}}}\left( {\left| {{Q_1}} \right| + \left| {{Q_2}} \right|} \right)\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}E = \frac{{4 \times 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}}}{{{{\left( {0.06{\rm{ m}}} \right)}^2}}} \times \left\{ {\left( {\left| {5.8 \times {{10}^{ - 6}}{\rm{ C}}} \right|} \right) + \left( {\left| { - 8 \times {{10}^{ - 6}}{\rm{ C}}} \right|} \right)} \right\}\\ = 1 \times {10^{13}}{\rm{ N/}}{{\rm{C}}^2} \times \left\{ {\left( {5.8 \times {{10}^{ - 6}}{\rm{ C}}} \right) + \left( {8 \times {{10}^{ - 6}}{\rm{ C}}} \right)} \right\}\\ = 1 \times {10^{13}}{\rm{ N/}}{{\rm{C}}^2} \times \left( {13.8 \times {{10}^{ - 6}}{\rm{ C}}} \right)\\ = 1.38 \times {10^8}{\rm{ N/C}}\end{aligned}\)

Thus, the magnitude of the electric field at a point midway between both charges is\(1.38 \times {10^8}{\rm{ N/C}}\).

The direction of an electric field of charge\({Q_1}\)is away from the positive charge. The direction of an electric field of charge\({Q_2}\)is toward the negative charge. The net electric field direction is toward the negative charge.

Thus, the direction of an electric field is toward the negative charge.