Four point charges are fixed in place as shown in the diagram.
a) What is the electric field experienced by the +2 µC charge?
b) What is the force on the +2 µC charge?
In this problem, you are first asked to find the electric field due to a grouping of point charges. This is a definition problem—the field is caused by the charges and you know an equation that relates them. Quantifying electric field is merely restating the charge and location information that you already have in a slightly different way. Likewise, electric force between point charges is also a definition problem for the same reasons.
Only the 2 µC charge is identified as the system of interest in this problem, because you want to know the field and force on that charge. The other charges are responsible for the interaction—they are the surroundings with which the system interacts.
This problem feels more complicated than many definition problems because the desired quantities are vectors. Therefore, you will need to do the additional math of dividing vectors into components. That math, however, does not change the overall approach to the problem.
Although many definition problems do not require pictures, you will want them in this case to help you with the vector addition that is required.
Part a):
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Part b):
Part a)
In equation form, electric field due to a point charge is defined as
E = (kq)/r^{2}
In this case, you will need the vector sum of the field due to three charges.
Part b)
Once you know electric field, the most straightforward way to work part b) is to use the definition of electric force on a charge in an electric field: F = qE.
If you are using this example for a problem in which you are only asked for the force due to point charges, you should use the definition of force due to a point charge
F = (k q_{1} q_{2})/r^{2}
and again you will need to find the vector sum of three forces.
Both methods are shown in this example.
Step 1
Find the electric field due to each point charge from the definition of electric field:
E_{-3} = (kq_{-3}) / r^{2}
E_{-3} = (9.0 x 10^{9} N∙m^{2}/C^{2})( 3 x 10^{-6} C) / (0.10 m)^{2}
E_{-3} = 2.7 x 10^{6} N/C
E_{-2} = (kq_{-2})/r^{2}
E_{-2} = (9.0 x 10^{9} N∙m^{2}/C^{2})( 2 x 10^{-6} C) / (0.20 m)^{2}
E_{-2} = 4.5 x 10^{5} N/C
E_{3} = (kq_{3})/r^{2}
E_{3} = (9.0 x 10^{9} N∙m^{2}/C^{2})( 3 x 10^{-6} C) / (0.224 m)^{2}
E_{3} = 5.4 x 10^{5} N/C
Find the x- and y-components of each field in preparation for doing the vector sum:
E_{3x} = E_{3}cos(27^{o}) = -5.4 x 10^{5} N/C cos(27^{o}) = -4.8 x 10^{5} N/C;
E_{3y} = E_{3}sin(27^{o}) = -5.4 x 10^{5} N/C sin(27^{o}) = -2.5 x 10^{5} N/C
E_{-3x} = 0; E_{-3y} = +2.7 x 10^{6} N/C
E_{-2x} = +4.5 x 10^{5} N/C; E_{-2}y = 0
Find the vector sum of the three fields to get the total field:
E_{x} = E_{-3x} + E_{-2x} + E_{3x} = 0 + 4.5 x 10^{5} N/C + (-4.8 x 10^{5} N/C) = -3 x 10^{4} N/C
E_{y} = E_{-3y} + E_{-2y} + E_{3y} = 2.7 x 10^{6} N/C + 0 + (-2.5 x 10^{5} N/C) = 2.5 x 10^{6} N/C
E^{2} = E_{x}^{2} + E_{y}^{2} = (-3 x 10^{4} N/C)^{2} + (2.5 x 10^{6} N/C)^{2} = 6.25 x 10^{12} N^{2}/C^{2}
E = 2.5 x 10^{6} N/C
tanθ = E_{y} / E_{x} = (2.5 x 10^{6} N/C) / (3 x 10^{4} N/C) = 83.3
θ = 89^{o}
The 2 µC charge experiences an electric field that has a magnitude of 2.5 x 10^{6} N/C and is directed at an angle of 89^{o} above the –x axis.
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Step 2
If you have already solved for electric field due to the point charges, as is the case in this problem, the definition relating electric field to electric force is all you need to find force:
The 2 µC charge experiences an electric force that has a magnitude of 5 N and is directed at an angle of 89^{o} above the –x axis.
If you don’t yet have the relation between force and field, you can find force beginning with the point charges.
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Step 3
Find the electric field on the 2µC charge due to each of the other point charges from the definition of electric force:
F_{2-3} = (kq_{-3}q_{2}) / r^{2}
F_{2-3} = (9.0 x 10^{9} N∙m^{2}/C^{2})( 3 x 10^{-6} C)(2 x 10^{-6} C) / (0.10 m)^{2}
F_{2-3} = 5.4 N
F_{2-2} = (kq_{-2}q_{2})/r^{2}
F_{2-2} = (9.0 x 10^{9} N∙m^{2}/C^{2})( 2 x 10^{-6} C)(2 x 10^{-6} C) / (0.20 m)^{2}
F_{2-2} = 0.9 N
F_{23} = (kq_{3})q_{2}/r^{2}
F_{23} = (9.0 x 10^{9} N∙m^{2}/C^{2})( 3 x 10^{-6} C)(2 x 10^{-6} C) / (0.224 m)^{2}
F_{23} = 1.1 N
Find the x- and y-components of each field in preparation for doing the vector sum:
F_{23x} = F_{23}cos(27^{o}) = -1.1 N cos(27^{o}) = -0.96 N;
F_{23y} = F_{23}sin(27^{o}) = -1.1 N sin(27^{o}) = -0.49 N
F_{2-3x} = 0; F_{2-3y} = +5.4 N
F_{2-2x} = +0.9 N; F_{2-2}y = 0
Find the vector sum of the three fields to get the total field:
F_{x} = F_{2-3x} + F_{2-2x} + F_{23x} = 0 + 0.9 N + (-0.06 N
F_{y} = F_{2-3y} + F_{2-2y} + F_{23y} = 5.4 N + 0 + (-0.49 N) = 4.9 N
F^{2} = F_{x}^{2} + F_{y}^{2} = (-0.06 N)^{2} + (4.9 N)^{2} = 24 N^{2}
F = 4.9 N
tanθ = F_{y} / F_{x} = (4.9 N) / (0.06 N) = 82
θ = 89^{o}
The 2 µC charge experiences an electric force that has a magnitude of 5 N and is directed at an angle of 89^{o} above the –x axis.
No. In Newton’s Second Law, you are asked to relate the forces on an object to the motion of that object. In this case, you aren’t asked about motion (in fact, the charges are fixed in place by non-specified forces) but rather you are just asked for the electric force and field. This is the equivalent of being asked for gravitational force when you are given masses.
Electric field is defined as having the same direction as force on a positive charge. In other words, electric field due to a negative charge points in towards the charge, and electric field due to a positive charge points out away from the charge. So to draw the electric field at any point, sketch a guide line between that point and the charge whose field you want to draw. Then draw the field in or out along that guideline.
An object cannot put a force on itself—it doesn’t feel its own field. So the 2 µC charge experiences a force due to (the field of) the -2 µC charge, the 3 µC charge, and the -3 µC charge, but not due to itself.
You probably learned this as “opposite angles are congruent” in geometry. When two lines intersect (in this case, the x-axis and the guideline between the origin and the 3 µC charge,) the angles opposite each other are the same.
Electric field is defined as having the same direction as force on a positive charge. In other words, electric field due to a negative charge points in towards the charge, and electric field due to a positive charge points out away from the charge. So to draw the electric field at any point, sketch a guide line between that point and the charge whose field you want to draw. Then draw the field in or out along that guideline.
You probably learned this as “opposite angles are congruent” in geometry. When two lines intersect (in this case, the x-axis and the guideline between the origin and the 3 µC charge,) the angles opposite each other are the same.
The Coulomb (C) or unit of charge is quite large. Therefore, charge is often given in smaller units. One micro-Coulomb (µC) is 10^{-6} C.
The definition of electric field gives the magnitude of the electric field only. Direction comes from your coordinate system and the understanding that electric field points out from negative charges and in to positive charges. Therefore, it is best to use absolute values only in the equation. If you put in the sign of the charge, it is easy to mistake that for the direction of the field.
The +3 µC charge is 0.224 m away from the location at which you want to know the field.
Electric field is out from a positive charge. In this case, out from the 3 µC charge means that at the location of the 2 µC charge the field is down and to the left. In other words, the x- and y- components are both negative.
No. For one thing, each charge is at a different distance away from the point of interest. For another, you need to take the vector sum of the fields or forces. There isn’t a correct short cut.
The Coulomb (C) or unit of charge is quite large. Therefore, charge is often given in smaller units. One micro-Coulomb (µC) is 10^{-6} C.
The definition of electric force gives the magnitude of the electric force only. Direction comes from your coordinate system and the understanding that electric field points out from negative charges and in to positive charges. Therefore, it is best to use absolute values only in the equation. If you put in the sign of the charge, it is easy to mistake that for the direction of the force.
The +3 µC charge is 0.224 m away from the location at which you want to know the field.
Electric field is out from a positive charge. In this case, out from the 3 µC charge means that at the location of the 2 µC charge the field is down and to the left. In other words, the x- and y- components are both negative.
No. For one thing, each charge is at a different distance away from the point of interest. For another, you need to take the vector sum of the fields or forces. There isn’t a correct short cut.
The direction of electric field is defined as being radially out from a positive charge and radially in towards a negative charge.
Point charges do not experience an electric field due to their own charge. Therefore, the field on the 2 µC charge is the vector sum of three fields—those due to the -2 µC charge, the 3 µC charge, and the -3 µC charge.
No. In Newton’s Second Law, you are asked to relate the forces on an object to the motion of that object. In this case, you aren’t asked about motion (in fact, the charges are fixed in place by non-specified forces) but rather you are just asked for the electric force and field. This is the equivalent of being asked for gravitational force when you are given masses.
Electric field is defined as having the same direction as force on a positive charge. In other words, electric field due to a negative charge points in towards the charge, and electric field due to a positive charge points out away from the charge. So to draw the electric field at any point, sketch a guide line between that point and the charge whose field you want to draw. Then draw the field in or out along that guideline.
Like charges are repelled from each other; negative and positive charges are attracted to each other. So just draw a guide line between the charges and sketch the force along that line either towards or away from the charge of interest (in this case, the 2 µC charge.)
The Coulomb (C) or unit of charge is quite large. Therefore, charge is often given in smaller units. One micro-Coulomb (µC) is 10^{-6} C.
An object cannot put a force on itself—it doesn’t feel its own field. So the 2 µC charge experiences a force due to (the field of) the -2 µC charge, the 3 µC charge, and the -3 µC charge, but not due to itself.
You probably learned this as “opposite angles are congruent” in geometry. When two lines intersect (in this case, the x-axis and the guideline between the origin and the 3 µC charge,) the angles opposite each other are the same.
No. For one thing, each charge is at a different distance away from the point of interest. For another, you need to take the vector sum of the fields or forces. There isn’t a correct short cut.
No, you aren’t asked for the total force on the charge, you are just asked for the total electric force on the charge.
I use E_{-3} to refer to the electric field due to the -3 µC charge. At the origin, the direction of this field is from the origin towards the negative charge.
I use E_{-2} to refer to the electric field due to the -2 µC charge. At the origin, the direction of this field is from the origin towards the negative charge.
I use E_{3} to refer to the electric field due to the +3 µC charge. At the origin, the direction of this field is from the origin away from the positive charge.
The dotted line is merely a guide to correctly show the direction of E_{3}.
tan θ = (opposite side)/(adjacent side) = (0.10)/(0.20) = 0.50
θ = tan^{-1} (0.50) [or inv tan (0.50) depending on your calculator] = 27^{0}
As you can see in the figure, the x- and y-components of a vector make up the sides of a right triangle. The vector itself forms the hypotenuse (h). The side of the triangle opposite the angle that you use is given by h sinθ and the side that touches the angle you use is given by h cosθ (soh cah toa)
In this case, the x-component is adjacent to the 27^{0}angle, and so is given by E_{3}cos(27^{0}). Likewise, the y-component is opposite to the 27^{0}angle and is therefore given by E_{3}sin(27^{0}).
You know this angle is 27^{0} because when two lines intersect (in this case, the x-axis and the guideline between the origin and the 3 µC charge,) the angles opposite to each other are the same.
Point charges do not experience an electric force due to their own charge. Therefore, the force on the 2 µC charge is the vector sum of three forces—those due to interactions with the -2 µC charge, the 3 µC charge, and the -3 µC charge.
The direction of electric force is determined by the location of the charges and whether it is attractive (interaction between a + and a – charge) or repulsive (interaction of like charges.)
I use F_{2-3} to refer to the electric force between the 2 µC charge and the -3 µC charge. This interaction is attractive and on a line between the two charges.
I use F_{2-3} to refer to the electric force between the 2 µC charge and the +3 µC charge. This interaction is repulsive and on a line between the two charges.
I use F_{2-2} to refer to the electric force between the + 2 µC charge and the -2 µC charge. This interaction is attractive and on a line between the two charges.
The dotted line is merely a guide to correctly show the direction of F_{2-3}.
tan θ = (opposite side)/(adjacent side) = (0.10)/(0.20) = 0.50 θ = tan^{-1} (0.50) [or inv tan (0.50 depending on your calculator) = 27^{0}
As you can see in the figure, the x- and y-components of a vector make up the sides of a right triangle. The vector itself forms the hypotenuse (h). The side of the triangle opposite the angle that you use is given by h sinθ and the side that touches the angle you use is given by h cosθ (soh cah toa)
In this case, the x-component is adjacent to the 27^{0}angle, and so is given by F_{23}cos(27^{0}). Likewise, the y-component is opposite to the 27^{0}angle and is therefore given by F_{23}sin(27^{0}).
You know this angle is 27^{0} because when two lines intersect (in this case, the x-axis and the guideline between the origin and the 3 µC charge,) the angles opposite to each other are the same.
k = 9.0 x 10^{9} N∙m^{2}/C^{2}
Some books use 1/(4πε_{0}) instead of k. In either case, the value is the same.
The definition of electric field gives the magnitude of the electric field only. Direction comes from your coordinate system and the understanding that electric field points out from negative charges and in to positive charges. Therefore, it is best to use absolute values only in the equation. If you put in the sign of the charge, it is easy to mistake that for the direction of the field.
The -3 µC charge is 0.10 m away from the location at which you want to know the field.
The -2 µC charge is 0.20 m away from the location at which you want to know the field.
The +3 µC charge is 0.224 m away from the location at which you want to know the field.
The -3 µC charge is 0.10 m away from the location at which you want to know the force.
The -2 µC charge is 0.20 m away from the location at which you want to know the force.
The +3 µC charge is 0.224 m away from the location at which you want to know the force.
Electric field is out from a positive charge. In this case, out from the 3 µC charge means that at the location of the 2 µC charge the field is down and to the left. In other words, the x- and y- components are both negative.
Electric field acts in to a negative charge. In this case, in to the -3 µC charge means that at the location of the 2 µC charge the field is straight up. In other words, there is no x-component and the y-component is positive.
Electric force is attractive between positive and negative charges. In this case, the attraction to the -2 µC charge pulls the 2 µC charge to the right. In other words, there is no y-component to this force and the x-component is positive.”
E^{2} = E_{x}^{2} + E_{y}^{2} is the Pythagorean Theorem applied to the electric field vector.
For the angle shown in the figure, the y-component of electric field is opposite the angle and the x-component is adjacent.
The definition of electric force gives the magnitude of the force only. Direction comes from your coordinate system and the understanding that electric force is attractive between + and – charges and repulsive between same charges. Therefore, it is best to use absolute values only in the equation. If you put in the sign of the charge, it is easy to mistake that for the direction of the force.
Electric force is repulsive between like charges. In this case, “repelled from the 3 µC charge” means that at the 2 µC charge is pushed down and to the left. In other words, the x- and y- components of force are both negative.
Electric force is attractive between positive and negative charges. In this case, the attraction to the -3 µC charge pulls the 2 µC charge the field straight up. In other words, there is no x-component to this force and and the y-component is positive.
F^{2} = F_{x}^{2} + F_{y}^{2} is the Pythagorean Theorem applied to the electric force vector.
Electric field acts in to a negative charge. In this case, in to the -2 µC charge means that at the location of the 2 µC charge the field is straight to the right. In other words, there is no y-component and the x-component is positive.
For the angle shown in the figure, the y-component of electric force is opposite the angle and the x-component is adjacent.
The key to understanding this problem is that both parts are merely definition problems. It feels much more complicated (and takes much more work) because the quantities defined are vector quantities, and you are asked to find the sum of several individual fields (or forces.) Note that almost all of the mathematical effort in this problem is in finding the vector sum. If you don’t remember how to work with vectors and need more of a refresher than is provided in this problem, click here for an example of dividing vectors into components and click here for an example working with the result.
The definition F_{E} = (k q_{1} q_{2})/r^{2} is analogous to Fg = (k m_{1} m_{2})/r^{2}. Whenever you are given a distribution of point charges, you can find the force on one of those charges by
Electric field (E) is analogous to gravitational field (g). If you know the electric field, you can find the force on any object with a charge q in that field just by the definition F_{E} = qE (analogous to F_{g} = mg.)
You can also calculate electric field directly from point charges. (If you want to see where the equation comes from, just look at the color coding in the equations above. Remember, red represents the system you care about and blue represents the effect of the surroundings. So electric field contains all information about the surrounding point charges that act on your system.) Whenever you are given a distribution of point charges, you can find the electric field of those charges by