Two point charges are placed on the x axis. Figure 1. The first charge, q1= 8.00nC,…
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Question “Two point charges are placed on the x axis. Figure 1. The first charge, q1= 8.00nC,…”
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
This concept is used to solve the problem of electric field.
First, use the electric field formula to calculate the electric fields at point O. The expression for electric fields will later be used to calculate the net electric force at the point O’s net electric field.
The force per unit of charge is the electric field. If the direction of the electric field is for the positive charge, it is radially outward. However, if the direction is for the negative charge, it is radially inward.
The electric field is:
E = \frac{{kq}}{{{r^2}}}E refers to the electric field, K the Coulomb’s constant and Q the charge. R represents the distance.
(a)
Refer to figure 1.
Consider two points charged placed on the positive and negative sides of the x-axis. Calculate the electric field caused by these two charges at origin.
The electric field $math_tag_1
{E_2} = \frac{{k{q_2}}}{{{r_2}^2}}\hat iThe electric field $math_tag_3
{E_1} = - \frac{{k{q_1}}}{{{r_1}^2}}\hat iThe net electric field $math_tag_5
E = {E_1} + {E_2}Substitute $math_tag_7
{E_x} = - \frac{{k{q_1}}}{{{r_1}^2}}\hat i + \frac{{k{q_2}}}{{{r_2}^2}}\hat iSubstitute $math_tag_9
\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}The y-axis does not contain an electric field component. The electric field in the direction of y is therefore zero.
The electric field at the y-axis therefore is.
{E_y} = 0The electric field $math_tag_12
E = \left( {0.385{\rm{ N/C}}} \right)\hat iThe electric field at the source is $math_tag_14
(b)
Refer to figure 2.
Consider two points charged placed on the positive and negative sides of the x-axis. Calculate the electric field caused by these two charges at origin.
The electric field $math_tag_1
{E_2} = - \frac{{k{q_2}}}{{{r_2}^2}}\hat iThe electric field $math_tag_3
{E_1} = - \frac{{k{q_1}}}{{{r_1}^2}}\hat iThe net electric field $math_tag_5
E = {E_1} + {E_2}Substitute $math_tag_7
{E_x} = - \frac{{k{q_1}}}{{{r_1}^2}}\hat i - \frac{{k{q_2}}}{{{r_2}^2}}\hat iSubstitute $math_tag_9
\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}The y-axis does not contain an electric field component. The electric field in the direction of y is therefore zero.
The electric field at the y-axis therefore is.
{E_y} = 0The electric field $math_tag_12
E = - \left( {0.948{\rm{ N/C}}} \right)\hat iThe direction of the electrical field at the source is toward the negative x-axis.
$math_tag_28 is the electric field at origin. Part a
Part of
Answer
The electric field at the source is $math_tag_14
Conclusion
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