Question

(II) The electric field midway between two equal but opposite point charges is\({\bf{386 N/C}}\)and the distance between the charges is 16.0 cm. What is the magnitude of the charge on each?

Short answer

The magnitude of each charge is \(1.37 \times {10^{ - 10}}{\rm{ C}}\).

Step-by-step solution

Understanding the electric field in the midway between two opposite charges

The two charges are placed at some distance.The magnitude of the electric field at a certain distance due to a point charge depends on the magnitude of the charge and distance from the center of the charge. The net electric field midway is the sum of the magnitudes of both electric fields.

Identification of given data

The given data can be listed below as:

• The distance between the two charges is\(d = 16{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) = 0.16{\rm{ m}}\).
• The electric field midway between the two charges is\(E = {\rm{386 N/C}}\).
• 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 midway is half the total distance (d/2). The charge \( + Q\) is positive and \( - Q\) is negative. \({\overrightarrow {\bf{E}} _{{\rm{ + Q}}}}\) and \({\overrightarrow {\bf{E}} _{ - {\rm{Q}}}}\) are the electric field vectors of charges \( + Q\) and\( - Q\).

Determination of the magnitude of each charge

The magnitude of an electric field of charge\( + Q\)can be expressed as:

\({E_{{\rm{ + Q}}}} = k\frac{{\left| { + Q} \right|}}{{{{\left( {\frac{d}{2}} \right)}^2}}}\) … (i)

The magnitude of an electric field of charge\( - Q\)can be expressed as:

\({E_{ - Q}} = k\frac{{\left| { - Q} \right|}}{{{{\left( {\frac{d}{2}} \right)}^2}}}\) … (ii)

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

\(\begin{aligned}{c}E = \left| {{E_{{\rm{ + Q}}}}} \right| + \left| {{E_{ - Q}}} \right|\\ = k\frac{{\left| { + Q} \right|}}{{{{\left( {\frac{d}{2}} \right)}^2}}} + k\frac{{\left| { - Q} \right|}}{{{{\left( {\frac{d}{2}} \right)}^2}}}\\ = 4k\frac{Q}{{{d^2}}} + 4k\frac{Q}{{{d^2}}}\\ = \frac{{4k}}{{{d^2}}} \times 2Q\end{aligned}\)

Further, solve the above equation.

\(\begin{aligned}{l}E = \frac{{8kQ}}{{{d^2}}}\\Q = \frac{{E{d^2}}}{{8k}}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}Q = \frac{{{\rm{386 N/C}} \times {{\left( {0.16{\rm{ m}}} \right)}^2}}}{{8 \times 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}}}\\ = \frac{{9.88}}{{7.2 \times {{10}^{10}}{\rm{ }}}}{\rm{ C}}\\ = 1.37 \times {10^{ - 10}}{\rm{ C}}\end{aligned}\)

Thus, the magnitude of each charge is \(1.37 \times {10^{ - 10}}{\rm{ C}}\)