Question

Two point charges are placed on the x axis.(Figure 1) The first charge, q1 = 8.00 nC , is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q2 = 6.00 nC , is placed a distance 9.00 m from the origin along the negative x axis.

Answer 1

Concepts and reason

The concept used to solve the problem are the electric field due to a point charge and superposition principle.

Use superposition principle and electric field expression due to a point charge to calculate electric field.

Fundamentals

Electric field is defined as the region around the charge where another charge experiences force.

Superposition principle says that electric force at desired point, is vector sum of electric force experienced by the point charge placed at that point, due to all point charges.

The electric field due to two-point charges is given by the following relation.

${\bf{E}} = \frac{{k{q_1}}}{{r_{1i}^2}}{{\bf{\hat r}}_{1i}} + \frac{{k{q_2}}}{{r_{2i}^2}}{{\bf{\hat r}}_{2i}}$

Here, ${\bf{E}}$ is total electric field, ${q_1}$ is magnitude of first charge, ${q_2}$ is magnitude of second charge, ${r_{1i}}$ is magnitude of vector joining charge ${q_1}$ and point $i$, ${{\bf{\hat r}}_{1i}}$ is unit vector of ${r_{1i}}$, ${r_{2i}}$ is magnitude of vector joining charge ${q_2}$ and point $i$, ${{\bf{\hat r}}_{2i}}$ is unit vector of ${r_{2i}}$ and $k$ is constant.

Part A

Since, ${q_1}$, ${q_2}$ are positive charge, electric field due to ${q_1}$ at origin will be towards negative x axis and due to ${q_2}$ it will be towards positive x axis.

$\begin{array}{c}\\{\bf{E}} = \frac{{k{q_1}}}{{r_{1i}^2}}\left( { - {\bf{\hat i}}} \right) + \frac{{k{q_2}}}{{r_{2i}^2}}{\rm{ }}{\bf{\hat i}}\\\\ = k\left( {\frac{{{q_1}}}{{r_{1i}^2}}\left( { - {\bf{\hat i}}} \right) + \frac{{{q_2}}}{{r_{2i}^2}}{\rm{ }}{\bf{\hat i}}} \right)\\\end{array}$

Substitute, $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}$ for $k$, $8.00{\rm{ nC}}$ for ${q_1}$ (charge of electron), $6.00{\rm{ nC}}$ for ${q_2}$, $16.0{\rm{ m}}$ for ${r_{1i}}$, and $9.00{\rm{ m}}$ for ${r_{2i}}$ (distance of respective charges from origin) in the above equation.

$\begin{array}{c}\\{\bf{E}} = \left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {\frac{{8.00{\rm{ nC}}}}{{{{\left( {16.0{\rm{ m}}} \right)}^2}}}\left( { - {\bf{\hat i}}} \right) + \frac{{6.00{\rm{ nC}}}}{{{{\left( {9.00{\rm{ m}}} \right)}^2}}}{\rm{ }}{\bf{\hat i}}} \right)\\\\ = \left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {0.043{\rm{ nC}} \cdot {{\rm{m}}^{ - 2}}} \right)\left( {\frac{{1{\rm{ C}}}}{{{{10}^9}{\rm{ nC}}}}} \right){\bf{\hat i}}\\\\ = 0.385{\rm{ N}}{{\rm{C}}^{ - 1}}{\rm{ }}{\bf{\hat i}}\\\end{array}$

Part B

Since, ${q_1}$, is positive charge, electric field due to ${q_1}$ at origin will be towards negative x axis and ${q_2}$ is negatively charged, electric field due to ${q_2}$ will be towards negative x axis.

$\begin{array}{c}\\{\bf{E}} = \frac{{k{q_1}}}{{r_{1i}^2}}\left( { - {\bf{\hat i}}} \right) + \frac{{k{q_2}}}{{r_{2i}^2}}\left( { - {\bf{\hat i}}} \right)\\\\ = k\left( {\frac{{{q_1}}}{{r_{1i}^2}} + \frac{{{q_2}}}{{r_{2i}^2}}} \right)\left( { - {\bf{\hat i}}} \right)\\\end{array}$

Substitute, $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}$ for $k$, $8.00{\rm{ nC}}$ for ${q_1}$ (charge of electron), $- 6.00{\rm{ nC}}$ for ${q_2}$, $16.0{\rm{ m}}$ for ${r_{1i}}$, and $9.00{\rm{ m}}$ for ${r_{2i}}$ (distance of respective charges from origin) in the above equation.

$\begin{array}{c}\\{\bf{E}} = \left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( { - \frac{{8.00{\rm{ nC}}}}{{{{\left( {16.0{\rm{ m}}} \right)}^2}}} + \frac{{ - 6.00{\rm{ nC}}}}{{{{\left( {9.00{\rm{ m}}} \right)}^2}}}} \right)\left( {{\bf{\hat i}}} \right)\\\\ = \left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {0.105{\rm{ nC}} \cdot {{\rm{m}}^{ - 2}}} \right)\left( {\frac{{1{\rm{ C}}}}{{{{10}^9}{\rm{ nC}}}}} \right)\left( { - {\bf{\hat i}}} \right)\\\\ = 0.948{\rm{ N}}{{\rm{C}}^{ - 1}}\left( { - {\bf{\hat i}}} \right)\\\end{array}$

Ans: Part A

Since there is no y component in the final result, the electric field in coordinate form, at origin is $\left( {0.385{\rm{ N}}{{\rm{C}}^{ - 1}},0{\rm{ N}}{{\rm{C}}^{ - 1}}} \right)$.

Part B

Since there is no y component in the final result, the electric field in coordinate form, at origin is $\left( { - 0.948{\rm{ N}}{{\rm{C}}^{ - 1}},0{\rm{ N}}{{\rm{C}}^{ - 1}}} \right)$.