Question

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

a) find the electric field at the origin, point O.    Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb.

b) assume that charge q2 is negative. q2= -6 nC (figure 2) what is the net electric field at the origin, point O? Give the x and y components of the electric field as an ordered pair. Express in newtons per coulomb.

figure 1

Figure 2

Answer 1

Concepts and reason

The concept is used to solve this problem is electric field.

Initially, calculate the electric field at origin, point O by using the electric field formula. Later calculate the net electric field at net electric field at point O by using the expression for electric field.

Fundamentals

The electric field is defined as the force per unit charge. The direction of electric field for the positive charge is radially outward or if the direction of electric field for the negative charge is radially inward.

The electric field is,

$E = \frac{{kq}}{{{r^2}}}$

Here, E is the electric field, k is the Coulomb’s constant, q is the charge, and r is the distance.

(a)

Refer figure 1 given in the question.

Consider two point charges placed on positive and negative x-axis respectively. Calculate the electric field due to these two charges at the origin.

The electric field ${E_2}$ due to charge ${q_2}$ placed on the negative x-axis at a distance of ${r_2}$ from the origin is,

${E_2} = \frac{{k{q_2}}}{{{r_2}^2}}\hat i$

The electric field ${E_1}$ due to charge ${q_1}$ placed on the positive x-axis at a distance of ${r_1}$ from the origin is,

${E_1} = - \frac{{k{q_1}}}{{{r_1}^2}}\hat i$

The net electric field $\left( E \right)$ due to both the charges on the x-axis is,

$E = {E_1} + {E_2}$

Substitute $- \frac{{k{q_1}}}{{{r_1}^2}}\hat i$ for ${E_1}$ and $\frac{{k{q_2}}}{{{r_2}^2}}\hat i$ for ${E_2}$ to find the electric field on the x-axis.

${E_x} = - \frac{{k{q_1}}}{{{r_1}^2}}\hat i + \frac{{k{q_2}}}{{{r_2}^2}}\hat i$

Substitute $8{\rm{ nC}}$ for ${q_2}$ , ${\rm{6 nC}}$ for ${q_1}$ , $- 9{\rm{ m}}$ for ${r_2}$ , $16{\rm{ m}}$ for ${r_1}$ , and $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}$ for k in the above equation.

$\begin{array}{c}\\{E_x} = - \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {8{\rm{ nC}}} \right)}}{{{{\left( {16{\rm{ m}}} \right)}^2}}}\left( {\frac{{{{10}^{ - 9}}{\rm{ C}}}}{{1{\rm{ nC}}}}} \right)\hat i + \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {{\rm{6 nC}}} \right)}}{{{{\left( { - 9{\rm{ m}}} \right)}^2}}}\left( {\frac{{{{10}^{ - 9}}{\rm{ C}}}}{{1{\rm{ nC}}}}} \right)\hat i\\\\{E_x} = \left( {0.385{\rm{ N/C}}} \right)\hat i\\\end{array}$

There is no electric field component in the y-axis. So, the electric field in the y-direction is zero.

Thus, the electric field in the y-axis is,

${E_y} = 0$

The electric field $E$ at the origin is due to the x-component of the electric field only.

$E = \left( {0.385{\rm{ N/C}}} \right)\hat i$

The electric field at the origin is $0.385{\rm{ N/C}}$ . The x-component and the y-component of the electric field is $\left( {0.385{\rm{ N/C}},0} \right)$ .

(b)

Refer figure 2 given in the question.

Consider two point charges placed on positive and negative x-axis respectively. Calculate the electric field due to these two charges at the origin.

The electric field ${E_2}$ due to charge ${q_2}$ placed on the negative x-axis at a distance of ${r_2}$ from the origin is,

${E_2} = - \frac{{k{q_2}}}{{{r_2}^2}}\hat i$

The electric field ${E_1}$ due to charge ${q_1}$ placed on the positive x-axis at a distance of ${r_1}$ from the origin is,

${E_1} = - \frac{{k{q_1}}}{{{r_1}^2}}\hat i$

The net electric field $\left( E \right)$ due to both the charges on the x-axis is,

$E = {E_1} + {E_2}$

Substitute $- \frac{{k{q_1}}}{{{r_1}^2}}\hat i$ for ${E_1}$ and $- \frac{{k{q_2}}}{{{r_2}^2}}\hat i$ for ${E_2}$ to find the electric field on the x-axis.

${E_x} = - \frac{{k{q_1}}}{{{r_1}^2}}\hat i - \frac{{k{q_2}}}{{{r_2}^2}}\hat i$

Substitute $8{\rm{ nC}}$ for ${q_2}$ , ${\rm{6 nC}}$ for ${q_1}$ , $- 9{\rm{ m}}$ for ${r_2}$ , $16{\rm{ m}}$ for ${r_1}$ , and $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}$ for k.

$\begin{array}{c}\\{E_x} = - \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {8{\rm{ nC}}} \right)}}{{{{\left( {16{\rm{ m}}} \right)}^2}}}\left( {\frac{{{{10}^{ - 9}}{\rm{ C}}}}{{1{\rm{ nC}}}}} \right)\hat i - \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {{\rm{6 nC}}} \right)}}{{{{\left( { - 9{\rm{ m}}} \right)}^2}}}\left( {\frac{{{{10}^{ - 9}}{\rm{ C}}}}{{1{\rm{ nC}}}}} \right)\hat i\\\\{E_x} = - \left( {0.948{\rm{ N/C}}} \right)\hat i\\\end{array}$

There is no electric field component in the y-axis. So, the electric field in the y-direction is zero.

Thus, the electric field in the y-axis is,

${E_y} = 0$

The electric field $E$ at the origin is due to the x-component of the electric field only.

$E = - \left( {0.948{\rm{ N/C}}} \right)\hat i$

The direction of the electric field at the origin is towards the negative x-axis.

The electric field at the origin is $- 0.948{\rm{ N/C}}$ . The x-component and the y-component of the electric field is $\left( { - 0.948{\rm{ N/C}},0} \right)$ .

Ans: Part a

Part a

Answer

The electric field at the origin is $0.385{\rm{ N/C}}$ . The x-component and the y-component of the electric field is $\left( {0.385{\rm{ N/C}},0} \right)$ .