03 Electrostatics Practice Problems
ELECTROSTATICS | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: A charge A of 1.7×10!! C is placed 2.0×10!! m from a charge B of 2.5×10!! C. What is the electric force being exerted on charge A from charge B.
PROBLEM 2: What is the electric field strength at a point where a 2.0×10!! C charge experiences an electric force of 5.30×10!! N?
PROBLEM 3: How much work must be done to bring a −7.0 µμC charged object to within 0.5 m of a 5.0 µμC charged object from a long way away? Is the work done negative or positive?
PROBLEM 4: The net force on the center charge is zero for the system of three collinear charges shown. What is the distance 𝑥? The charges have a magnitude of 2C, 3C, and 5C, respectively as shown below. 13|PREPINEER.COM
psst… don’t forget to take notesÎ
PROBLEM 5: Determine the magnitude of the electric field necessary to place a 1 N force on an electron.
14|PREPINEER.COM
psst… don’t forget to take notesÎ
ELECTROSTATICS | SOLUTIONS SOLUTION 1: In this problem we are looking for the forced exerted on charge from another based on the given strength of each charge, and the distance between the two charges. In the problem we are given: • A charge, 𝑄! = 1.70×10!! C • A charge, 𝑄! = 2.5×10!! C • The distance, 𝑟 = 2.0×10!! m • 𝑎!!" is assumed to be 1 • 𝜀 = 8.85×10!!" F m, which is the permittivity for free space The value for the permittivity for free space can be referenced on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Using Coulomb’s Law, we plug in the given values and solve for the force.
𝐹! =
𝑄! 𝑄! 𝑎 4𝜋𝜀𝑟 ! !"#
𝐹!→!
(1.7×10!! C)(2.5×10!! C) = 𝑎 4𝜋(8.85×10!!" F/m)(2.0×10!! m)! !!"
𝐹!→!
4.25×10!!" C = (1) 4.45×10!!" F/m
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psst… don’t forget to take notesÎ
𝐹!→! = 95.51 N
The formula for COULOMB’S LAW can be referenced on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
SOLUTION 2: In this problem we are solving for the strength of the electric field given the force exerted on a charge, with a given charge in Coulombs. In the problem we are given: • Force, 𝐹! = 5.30×10!! N • Charge, 𝑄 = 2.0×10!! C Since we know two of the three variables in the formula for electric field strength, we can solve for the missing intensity of the electric field.
𝐸=
𝐹! 𝑄
This formula is NOT provided in the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Plugging in our given values and solve for the electric strength we get
16|PREPINEER.COM
psst… don’t forget to take notesÎ
5.30×10!! N 𝐸= 2.0×10!! C 𝐸 = 265 N/C
SOLUTION 3: In this problem we are looking to calculate the amount of work it will take to bring a negatively charged object to a positively charged object, given a certain distance the negatively charged object must travel. In the problem we are given: • A charge, 𝑄! = −7.0×10!! C • A charge, 𝑄! = 5.0×10!! C • The distance, 𝑟 = 0.5 m • Coulomb’s Constant, 𝑘 = 9.0×10! Nm! /C ! We use the formula to calculate the work done, as we have all of the variables needed to solve for the work.
𝑊=
𝑄! 𝑄! 1 1 − 4𝜋𝜀 𝑟! 𝑟!
𝑊=
𝑘𝑄! 𝑄! 𝑟
Plugging in the variables and simplifying we calculate the work done. 17|PREPINEER.COM
psst… don’t forget to take notesÎ
9.0×10! Nm! /C ! −7.0×10!! C 5.0×10!! C 𝑊= 0.5 m 𝑊 = −0.63 J
Note: WORK is NEGATIVE because we are bringing a negative charge near a positive charge releasing energy therefore work is NEGATIVE. The formulas for ELECTROSTATIC FIELDS and COULOMB’S LAW can be referenced under the topic of ELECTROSTATIC FIELDS on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. However, this exact formula is NOT referenced in the Reference Handbook in this form.
SOLUTION 4: This is considered a “hard” problem you could face regarding electrostatics on the FE exam. This problem requires you to realize that the Coulomb Forces are in equilibrium, thus the net force on the center charge is zero. Equilibrium is defined as opposing forces balancing each other, resulting in a net resultant force of zero.
18|PREPINEER.COM
psst… don’t forget to take notesÎ
As in general statics and other physics problem we can write about the center charge, and set the equation equal to zero since we are told the net force on the center charge is zero. 𝐹!" − 𝐹!" = 0
We can then add the 𝐹!" to the right hand side of the equation and set the forces equal to each other.
𝐹!" = 𝐹!"
Since we know the magnitude of the charge on each charge, and have a distance in terms of 𝑥, we can plug these values into our formula for Coulomb’s Law to calculate the Coulomb Force on each side of the center charge.
𝐹! =
𝑄! 𝑄! 𝑎 4𝜋𝜀𝑟 ! !!"
The formula for COULOMB’S LAW can be referenced on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. The tricky part of this problem is realizing that the distance between the first and third charge is 2 m, so we will need to write the distance in terms of the variable 𝑥. 19|PREPINEER.COM
psst… don’t forget to take notesÎ
𝑟!" = 2 𝑟!" = 2 − 𝑥
For the distance between the first charge and center charge we subtract the total distance of 2 m between the first and third charge, by the given distance between the center and third charge of 𝑥. 𝑟!" = 2 − 𝑥 𝑟!" = 𝑥
We define the forces 𝐹!" and 𝐹!" in terms of Coulomb’s Law.
𝐹!" =
𝑄! 𝑄! 𝑎 4𝜋𝜀𝑟 ! !!"
𝐹!" =
𝑄! 𝑄! 𝑎 4𝜋𝜀𝑟 ! !!"
We then set the equations equal to each other in terms of Coulomb’s Law. 𝑄! 𝑄! 𝑄! 𝑄! = 4𝜋𝜀𝑟 ! 4𝜋𝜀𝑟 ! 𝑄! 𝑄! 4𝜋𝜀 2 − 𝑥
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!
=
𝑄! 𝑄! 4𝜋𝜀 𝑥 !
psst… don’t forget to take notesÎ
It is important to remember that even though we are not given a number value for the distance, we can set up an equation to solve for x since there is only one unknown variable in the equation.
By cross multiplying both sides of the equation we get
𝑄! 𝑄! 𝑥 ! = 𝑄! 𝑄! 2 − 𝑥
!
Plugging in our given values, the equation becomes
2 3 𝑥! = 3 5 2 − 𝑥
!
We Simplify the expression and identify the equation in terms of the variable 𝑥. 6𝑥 ! − 60 + 60𝑥 − 15𝑥 ! = 0 We then set up the equation in quadratic form so we can utilize the quadratic equation. 9𝑥 ! − 60𝑥 + 60 = 0 Identify the variables in the quadratic expression and plug them into the quadratic equation. Quadratic form: 21|PREPINEER.COM
psst… don’t forget to take notesÎ
𝑎𝑥 ! − 𝑏𝑥 + 𝑐 = 0 Quadratic Formula, −𝑏 ± 𝑏 ! − 4𝑎𝑐 𝑥= 2𝑎 Where 𝑎=9 𝑏 = −60 𝑐 = 60
Plugging in the values of 𝑎, 𝑏, and 𝑐, solve for 𝑥.
𝑥=
− −60 ±
−60 2 9
!
− 4 9 60
𝑥 = 5.44, 1.23
As a quadratic has two roots, we must determine which value of 𝑥 satisfies our problem. Since we know that the full distance is 2 m, the answer must be 𝑥 = 1.23 m as 𝑥 = 5.44 m would not possible in the problem.
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psst… don’t forget to take notesÎ
The formula for THE QUADRATIC EQUATION can be referenced on under the topic of MATHEMATICS on page 21 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. NOTE: Its recommended that you use the solve feature on your calculator. Once we had all the numbers plugged in on both sides of the equation, we could have simply used the solve function to solve for the values of 𝑥. This would have saved us the time to cross multiply, simplify, and use the quadratic equation.
SOLUTION 5: This problem requires you to calculate the magnitude of an electric field given the force exerted on the charge, and the magnitude of the charge. The hardest part about this problem is realizing that the magnitude of the charge is not directly given. By stating the problem, “force on the electron”, it is implied that the charge is the charge for one electron. In the problem we are given: • The charge for one electron, 𝑄!"!#$%&' = 1.602×10!!" C • Force, 𝐹! = 1 N We use the formula for the strength of an electric field related by force and charge.
𝐹! =
𝑄!"!#$%&' 𝑎 4𝜋𝜀𝑟 ! !!"
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psst… don’t forget to take notesÎ
𝐸=
𝐹! 𝑄!"!#$%&'
We then plug in our values for the force on the charge and the magnitude of the charge in Coulombs.
𝐸= 𝐸=
𝐹! 𝑄!"!#$%&' 1 N 1.602×10!!" C
We then solve for the magnitude of the electric field. 𝐸 = 6.24×10!" N/C 𝐸 = 6.24×10!" V/m
The formula for COULOMB’S LAW can be referenced on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
24|PREPINEER.COM
psst… don’t forget to take notesÎ
PROBLEM 1: A charge A of 1.7×10!! C is placed 2.0×10!! m from a charge B of 2.5×10!! C. What is the electric force being exerted on charge A from charge B.
PROBLEM 2: What is the electric field strength at a point where a 2.0×10!! C charge experiences an electric force of 5.30×10!! N?
PROBLEM 3: How much work must be done to bring a −7.0 µμC charged object to within 0.5 m of a 5.0 µμC charged object from a long way away? Is the work done negative or positive?
PROBLEM 4: The net force on the center charge is zero for the system of three collinear charges shown. What is the distance 𝑥? The charges have a magnitude of 2C, 3C, and 5C, respectively as shown below. 13|PREPINEER.COM
psst… don’t forget to take notesÎ
PROBLEM 5: Determine the magnitude of the electric field necessary to place a 1 N force on an electron.
14|PREPINEER.COM
psst… don’t forget to take notesÎ
ELECTROSTATICS | SOLUTIONS SOLUTION 1: In this problem we are looking for the forced exerted on charge from another based on the given strength of each charge, and the distance between the two charges. In the problem we are given: • A charge, 𝑄! = 1.70×10!! C • A charge, 𝑄! = 2.5×10!! C • The distance, 𝑟 = 2.0×10!! m • 𝑎!!" is assumed to be 1 • 𝜀 = 8.85×10!!" F m, which is the permittivity for free space The value for the permittivity for free space can be referenced on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Using Coulomb’s Law, we plug in the given values and solve for the force.
𝐹! =
𝑄! 𝑄! 𝑎 4𝜋𝜀𝑟 ! !"#
𝐹!→!
(1.7×10!! C)(2.5×10!! C) = 𝑎 4𝜋(8.85×10!!" F/m)(2.0×10!! m)! !!"
𝐹!→!
4.25×10!!" C = (1) 4.45×10!!" F/m
15|PREPINEER.COM
psst… don’t forget to take notesÎ
𝐹!→! = 95.51 N
The formula for COULOMB’S LAW can be referenced on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
SOLUTION 2: In this problem we are solving for the strength of the electric field given the force exerted on a charge, with a given charge in Coulombs. In the problem we are given: • Force, 𝐹! = 5.30×10!! N • Charge, 𝑄 = 2.0×10!! C Since we know two of the three variables in the formula for electric field strength, we can solve for the missing intensity of the electric field.
𝐸=
𝐹! 𝑄
This formula is NOT provided in the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Plugging in our given values and solve for the electric strength we get
16|PREPINEER.COM
psst… don’t forget to take notesÎ
5.30×10!! N 𝐸= 2.0×10!! C 𝐸 = 265 N/C
SOLUTION 3: In this problem we are looking to calculate the amount of work it will take to bring a negatively charged object to a positively charged object, given a certain distance the negatively charged object must travel. In the problem we are given: • A charge, 𝑄! = −7.0×10!! C • A charge, 𝑄! = 5.0×10!! C • The distance, 𝑟 = 0.5 m • Coulomb’s Constant, 𝑘 = 9.0×10! Nm! /C ! We use the formula to calculate the work done, as we have all of the variables needed to solve for the work.
𝑊=
𝑄! 𝑄! 1 1 − 4𝜋𝜀 𝑟! 𝑟!
𝑊=
𝑘𝑄! 𝑄! 𝑟
Plugging in the variables and simplifying we calculate the work done. 17|PREPINEER.COM
psst… don’t forget to take notesÎ
9.0×10! Nm! /C ! −7.0×10!! C 5.0×10!! C 𝑊= 0.5 m 𝑊 = −0.63 J
Note: WORK is NEGATIVE because we are bringing a negative charge near a positive charge releasing energy therefore work is NEGATIVE. The formulas for ELECTROSTATIC FIELDS and COULOMB’S LAW can be referenced under the topic of ELECTROSTATIC FIELDS on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. However, this exact formula is NOT referenced in the Reference Handbook in this form.
SOLUTION 4: This is considered a “hard” problem you could face regarding electrostatics on the FE exam. This problem requires you to realize that the Coulomb Forces are in equilibrium, thus the net force on the center charge is zero. Equilibrium is defined as opposing forces balancing each other, resulting in a net resultant force of zero.
18|PREPINEER.COM
psst… don’t forget to take notesÎ
As in general statics and other physics problem we can write about the center charge, and set the equation equal to zero since we are told the net force on the center charge is zero. 𝐹!" − 𝐹!" = 0
We can then add the 𝐹!" to the right hand side of the equation and set the forces equal to each other.
𝐹!" = 𝐹!"
Since we know the magnitude of the charge on each charge, and have a distance in terms of 𝑥, we can plug these values into our formula for Coulomb’s Law to calculate the Coulomb Force on each side of the center charge.
𝐹! =
𝑄! 𝑄! 𝑎 4𝜋𝜀𝑟 ! !!"
The formula for COULOMB’S LAW can be referenced on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. The tricky part of this problem is realizing that the distance between the first and third charge is 2 m, so we will need to write the distance in terms of the variable 𝑥. 19|PREPINEER.COM
psst… don’t forget to take notesÎ
𝑟!" = 2 𝑟!" = 2 − 𝑥
For the distance between the first charge and center charge we subtract the total distance of 2 m between the first and third charge, by the given distance between the center and third charge of 𝑥. 𝑟!" = 2 − 𝑥 𝑟!" = 𝑥
We define the forces 𝐹!" and 𝐹!" in terms of Coulomb’s Law.
𝐹!" =
𝑄! 𝑄! 𝑎 4𝜋𝜀𝑟 ! !!"
𝐹!" =
𝑄! 𝑄! 𝑎 4𝜋𝜀𝑟 ! !!"
We then set the equations equal to each other in terms of Coulomb’s Law. 𝑄! 𝑄! 𝑄! 𝑄! = 4𝜋𝜀𝑟 ! 4𝜋𝜀𝑟 ! 𝑄! 𝑄! 4𝜋𝜀 2 − 𝑥
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!
=
𝑄! 𝑄! 4𝜋𝜀 𝑥 !
psst… don’t forget to take notesÎ
It is important to remember that even though we are not given a number value for the distance, we can set up an equation to solve for x since there is only one unknown variable in the equation.
By cross multiplying both sides of the equation we get
𝑄! 𝑄! 𝑥 ! = 𝑄! 𝑄! 2 − 𝑥
!
Plugging in our given values, the equation becomes
2 3 𝑥! = 3 5 2 − 𝑥
!
We Simplify the expression and identify the equation in terms of the variable 𝑥. 6𝑥 ! − 60 + 60𝑥 − 15𝑥 ! = 0 We then set up the equation in quadratic form so we can utilize the quadratic equation. 9𝑥 ! − 60𝑥 + 60 = 0 Identify the variables in the quadratic expression and plug them into the quadratic equation. Quadratic form: 21|PREPINEER.COM
psst… don’t forget to take notesÎ
𝑎𝑥 ! − 𝑏𝑥 + 𝑐 = 0 Quadratic Formula, −𝑏 ± 𝑏 ! − 4𝑎𝑐 𝑥= 2𝑎 Where 𝑎=9 𝑏 = −60 𝑐 = 60
Plugging in the values of 𝑎, 𝑏, and 𝑐, solve for 𝑥.
𝑥=
− −60 ±
−60 2 9
!
− 4 9 60
𝑥 = 5.44, 1.23
As a quadratic has two roots, we must determine which value of 𝑥 satisfies our problem. Since we know that the full distance is 2 m, the answer must be 𝑥 = 1.23 m as 𝑥 = 5.44 m would not possible in the problem.
22|PREPINEER.COM
psst… don’t forget to take notesÎ
The formula for THE QUADRATIC EQUATION can be referenced on under the topic of MATHEMATICS on page 21 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. NOTE: Its recommended that you use the solve feature on your calculator. Once we had all the numbers plugged in on both sides of the equation, we could have simply used the solve function to solve for the values of 𝑥. This would have saved us the time to cross multiply, simplify, and use the quadratic equation.
SOLUTION 5: This problem requires you to calculate the magnitude of an electric field given the force exerted on the charge, and the magnitude of the charge. The hardest part about this problem is realizing that the magnitude of the charge is not directly given. By stating the problem, “force on the electron”, it is implied that the charge is the charge for one electron. In the problem we are given: • The charge for one electron, 𝑄!"!#$%&' = 1.602×10!!" C • Force, 𝐹! = 1 N We use the formula for the strength of an electric field related by force and charge.
𝐹! =
𝑄!"!#$%&' 𝑎 4𝜋𝜀𝑟 ! !!"
23|PREPINEER.COM
psst… don’t forget to take notesÎ
𝐸=
𝐹! 𝑄!"!#$%&'
We then plug in our values for the force on the charge and the magnitude of the charge in Coulombs.
𝐸= 𝐸=
𝐹! 𝑄!"!#$%&' 1 N 1.602×10!!" C
We then solve for the magnitude of the electric field. 𝐸 = 6.24×10!" N/C 𝐸 = 6.24×10!" V/m
The formula for COULOMB’S LAW can be referenced on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
24|PREPINEER.COM
psst… don’t forget to take notesÎ
