03 Electrostatics Review
ELECTROSTATICS | CONCEPT OVERVIEW The topic of ELECTROSTATICS can be referenced on page 199 of the NCEES Supplied Reference Handbook, 9.3 version for Computer Based Testing. Electrical circuit behavior is a result of a force, which has been found to exist between two charges called the COULOMB FORCE. If the charges have the same sign, the force is REPULSIVE. If the charges have opposite signs, the force is ATTRACTIVE. ELECTRIC CHARGE is given in units of Coulombs (C) where 1 C = 6.26×10!" Electrons or 1 Electron = 1.602×10!!" C The conversion for COULOMB TO ELECTRONS can be referenced under the TABLE OF FUNDAMENTAL CONSTANTS on page 1 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Charge is ALWAYS either POSITIVE + or NEGATIVE − . CONDUCTORS are materials in which charges can move freely. Metals are the best conductors.
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Coulomb’s Law is the formula in electrostatics that relates COULOMB FORCES and DISTANCE between point charges. It is expressed as
𝐹! =
𝑄! 𝑄! 𝑎 4𝜋𝜀𝑟 ! !!"
where • 𝐹! = the force on charge 2 due to charge 1 • 𝑄! = the ith point charge • 𝑟 = the distance between charges 1 and 2 • 𝑎!!" = a unit vector directed from 1 to 2 (For the FE, assume this will be 1) • 𝜀 = the permittivity of the medium For free space or air, 𝜀 = 𝜀! = 8.85×10!!" F m The formula for COULOMB’S LAW and for the permittivity of free space can both be referenced on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
CONCEPT EXAMPLE: Two points charges of 1.8×10!! C and 2.4×10!! C produce a force of 2.2×10!! N on each other. How far apart are these two charges?
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psst… don’t forget to take notesÎ
SOLUTION: In this problem we will be using electrostatics and Coulomb’s Law to determine the distance between two point charges based on the force they exert on each other. In the problem we are given: • Charge 1, 𝑄! = 1.8×10!! C • Charge 2, 𝑄! = 2.4×10!! C • Force, 𝐹 = 2.2×10!! N • 𝑎!!" is assumed to be 1 • 𝜀 = 8.85×10!!" F m Using the formula for Coulomb’s Law we will solve for the distance.
𝐹! =
𝑄! 𝑄! 𝑎 4𝜋𝜀𝑟 ! !!"
The formula for COULOMB’S LAW and for the permittivity of free space can both be referenced on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Plugging in the values for Charge 1, Charge 2, the Force, and unit vector we get
𝐹! =
1.8×10!! C 2.4×10!! C 4𝜋 8.85×10!!" F m 𝑟 !
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psst… don’t forget to take notesÎ
𝐹! =
0.0388 𝑟!
2.2×10!! 𝑁 =
0.0388 𝑟!
Simplifying the terms, we are left with one variable, 𝑟. We cross multiply so we can isolate the variable r and solve for the value. Solving for 𝑟, we get the distance between the two point charges. 𝑟 ! 2.2×10!! 𝑁 = 0.0388
𝑟=
0.0388 2.2×10!! 𝑁
𝑟 = 4.19 𝑚
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psst… don’t forget to take notesÎ
ELECTRIC FIELD An ELECTRIC FIELD represents the effects of all the charges in a system. We can describe the electric force on a charge q as the result of an ELECTRIC FIELD, E. An electric field exists at any point in space where a test charge, if placed at that point, would experience an ELECTRICAL FORCE. The STRENGTH of electric field depends on the size of and distance to the test charge.
The ELECTRIC FIELD INTENSITY at point 2 due to a point charge 𝑄! at point 1 can be expressed as
𝐸=
𝑄! 𝑎 4𝜋𝜀𝑟 ! !!" !
𝐹 = 𝑄𝐸 or 𝐹 = ! where • 𝐸 = the Electric field intensity at point 2 due to a point charge at p • 𝐹 = the electrostatic force in Newtons, N • 𝑄 = test charge in Coulombs, C 5|PREPINEER.COM
psst… don’t forget to take notesÎ
The formulas for ELECTROSTATIC FIELDS 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. Substituting this equation into Coulomb’s Law, the electric field intensity can be expressed as
𝐸=
𝑘𝑄 𝑟!
where • 𝑘 = Coulomb’s constant, 𝑘 = 9.0×10! Nm! C ! • 𝑟 = the distance between charges in meters, m This formula is NOT provided in the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
CONCEPT EXAMPLE An electric field has a strength of 6000 N C, at a distance of 1.5 m. What is the strength of the field at a distance of 6.0 m?
SOLUTION: In this problem we are solving for the strength of the electric field at a certain distance, based on the electric field at a given distance. 6|PREPINEER.COM
psst… don’t forget to take notesÎ
Using the formula for the electric field intensity, we will solve for the charge at the first distance as we are given the initial field strength and initial distance, but not the charge.
𝐸=
𝑘𝑄 𝑟!
This formula is NOT provided in the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Solving for the charge 𝑄, and plugging in the electric field strength, distance, and Coulomb’s constant, we find that 𝐸𝑟 ! 𝑄= 𝑘 6000 N 𝐶 1.5 m 𝑄= 9.0×10! Nm! C !
!
𝑄 = 1.5×10!! C
Now that we have the charge, we can use the same charge to calculate the electric field strength at the second specified distance.
𝐸! =
𝑘𝑄 𝑟!!
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psst… don’t forget to take notesÎ
This formula is NOT provided in the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. (9.0×10! Nm! /C ! )(1.5×10!! C) 𝐸! = 6.0 m ! 𝐸! = 375 N/C
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psst… don’t forget to take notesÎ
ELECTRIC POTENTIAL (VOLTAGE) The POTENTIAL DIFFERENCE or VOLTAGE from one point, A, to another point, B, is the WORK done against electrical forces in carrying a unit positive test charge from A to B. We represent the potential difference from A to B by 𝑉! − 𝑉! or by 𝛥𝑉. Its units are those of work per charge (Joules/Coulomb), called VOLTS (V).
WORK is done when charges are forced to move through a load by a SOURCE. The work W done in transporting a charge q from one point, A, to a second point, B, parallel to the E field is given in units of Joules (J). This work appears as ELECTRICAL POTENTIAL ENERGY. 𝑊 = 𝑞 𝑉! − 𝑉! = 𝑞𝛥𝑉 This formula is NOT provided in the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. 9|PREPINEER.COM
psst… don’t forget to take notesÎ
It is important to remember that bringing a positive charge near another positive charge requires INPUT therefore, the work is POSITIVE. Bringing a negative charge, near a positive charge, RELEASES energy therefore, the work is NEGATIVE. The work W performed on a moving charge, 𝑄! a certain distance in a field created by charge, 𝑄! is represented with the electrostatic field formula in the form of Coulomb’s Law.
𝑊=
𝑄! 𝑄! 1 1 − 4𝜋𝜀 𝑟! 𝑟!
𝑊=
𝑘𝑄! 𝑄! 𝑟
where • 𝑄! = the charge at point A in Coulombs, C • 𝑄! = the charge at point B in Coulombs, C • 𝜀 = 8.85×10!!" F/m • 𝑘 = 9.0×10! Nm! /C ! = Coulomb’s Constant 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.
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psst… don’t forget to take notesÎ
CONCEPT EXAMPLE: How much work must be done to bring a 4.0 𝜇C charged object to within 1.0 m of a 6.0 𝜇C charged object from a long way away? Is the work positive or negative?
SOLUTION: In this problem we will solve for the amount of work needed to move a charge based on electrostatics. In the problem, we are given: • A charge, 𝑄! = 4.0×10!! C • A charge, 𝑄! = 6.0×10!! C • The distance between the charges, 𝑟 = 1.0 m
Plugging these values into the formula for work and solving for W
𝑊=
𝑘𝑄! 𝑄! 𝑟
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(9.0×10! Nm! /C ! )(4.0×10!! C)(6.0×10!! C) 𝑊= 1.0 m 𝑊 = 0.216 J
Work is POSITIVE because we are bringing a positive charge near another positive charge requiring input, therefore the work is positive. 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.
12|PREPINEER.COM
psst… don’t forget to take notesÎ
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psst… don’t forget to take notesÎ
Coulomb’s Law is the formula in electrostatics that relates COULOMB FORCES and DISTANCE between point charges. It is expressed as
𝐹! =
𝑄! 𝑄! 𝑎 4𝜋𝜀𝑟 ! !!"
where • 𝐹! = the force on charge 2 due to charge 1 • 𝑄! = the ith point charge • 𝑟 = the distance between charges 1 and 2 • 𝑎!!" = a unit vector directed from 1 to 2 (For the FE, assume this will be 1) • 𝜀 = the permittivity of the medium For free space or air, 𝜀 = 𝜀! = 8.85×10!!" F m The formula for COULOMB’S LAW and for the permittivity of free space can both be referenced on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
CONCEPT EXAMPLE: Two points charges of 1.8×10!! C and 2.4×10!! C produce a force of 2.2×10!! N on each other. How far apart are these two charges?
2|PREPINEER.COM
psst… don’t forget to take notesÎ
SOLUTION: In this problem we will be using electrostatics and Coulomb’s Law to determine the distance between two point charges based on the force they exert on each other. In the problem we are given: • Charge 1, 𝑄! = 1.8×10!! C • Charge 2, 𝑄! = 2.4×10!! C • Force, 𝐹 = 2.2×10!! N • 𝑎!!" is assumed to be 1 • 𝜀 = 8.85×10!!" F m Using the formula for Coulomb’s Law we will solve for the distance.
𝐹! =
𝑄! 𝑄! 𝑎 4𝜋𝜀𝑟 ! !!"
The formula for COULOMB’S LAW and for the permittivity of free space can both be referenced on page 199 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Plugging in the values for Charge 1, Charge 2, the Force, and unit vector we get
𝐹! =
1.8×10!! C 2.4×10!! C 4𝜋 8.85×10!!" F m 𝑟 !
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1
psst… don’t forget to take notesÎ
𝐹! =
0.0388 𝑟!
2.2×10!! 𝑁 =
0.0388 𝑟!
Simplifying the terms, we are left with one variable, 𝑟. We cross multiply so we can isolate the variable r and solve for the value. Solving for 𝑟, we get the distance between the two point charges. 𝑟 ! 2.2×10!! 𝑁 = 0.0388
𝑟=
0.0388 2.2×10!! 𝑁
𝑟 = 4.19 𝑚
4|PREPINEER.COM
psst… don’t forget to take notesÎ
ELECTRIC FIELD An ELECTRIC FIELD represents the effects of all the charges in a system. We can describe the electric force on a charge q as the result of an ELECTRIC FIELD, E. An electric field exists at any point in space where a test charge, if placed at that point, would experience an ELECTRICAL FORCE. The STRENGTH of electric field depends on the size of and distance to the test charge.
The ELECTRIC FIELD INTENSITY at point 2 due to a point charge 𝑄! at point 1 can be expressed as
𝐸=
𝑄! 𝑎 4𝜋𝜀𝑟 ! !!" !
𝐹 = 𝑄𝐸 or 𝐹 = ! where • 𝐸 = the Electric field intensity at point 2 due to a point charge at p • 𝐹 = the electrostatic force in Newtons, N • 𝑄 = test charge in Coulombs, C 5|PREPINEER.COM
psst… don’t forget to take notesÎ
The formulas for ELECTROSTATIC FIELDS 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. Substituting this equation into Coulomb’s Law, the electric field intensity can be expressed as
𝐸=
𝑘𝑄 𝑟!
where • 𝑘 = Coulomb’s constant, 𝑘 = 9.0×10! Nm! C ! • 𝑟 = the distance between charges in meters, m This formula is NOT provided in the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing.
CONCEPT EXAMPLE An electric field has a strength of 6000 N C, at a distance of 1.5 m. What is the strength of the field at a distance of 6.0 m?
SOLUTION: In this problem we are solving for the strength of the electric field at a certain distance, based on the electric field at a given distance. 6|PREPINEER.COM
psst… don’t forget to take notesÎ
Using the formula for the electric field intensity, we will solve for the charge at the first distance as we are given the initial field strength and initial distance, but not the charge.
𝐸=
𝑘𝑄 𝑟!
This formula is NOT provided in the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Solving for the charge 𝑄, and plugging in the electric field strength, distance, and Coulomb’s constant, we find that 𝐸𝑟 ! 𝑄= 𝑘 6000 N 𝐶 1.5 m 𝑄= 9.0×10! Nm! C !
!
𝑄 = 1.5×10!! C
Now that we have the charge, we can use the same charge to calculate the electric field strength at the second specified distance.
𝐸! =
𝑘𝑄 𝑟!!
7|PREPINEER.COM
psst… don’t forget to take notesÎ
This formula is NOT provided in the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. (9.0×10! Nm! /C ! )(1.5×10!! C) 𝐸! = 6.0 m ! 𝐸! = 375 N/C
8|PREPINEER.COM
psst… don’t forget to take notesÎ
ELECTRIC POTENTIAL (VOLTAGE) The POTENTIAL DIFFERENCE or VOLTAGE from one point, A, to another point, B, is the WORK done against electrical forces in carrying a unit positive test charge from A to B. We represent the potential difference from A to B by 𝑉! − 𝑉! or by 𝛥𝑉. Its units are those of work per charge (Joules/Coulomb), called VOLTS (V).
WORK is done when charges are forced to move through a load by a SOURCE. The work W done in transporting a charge q from one point, A, to a second point, B, parallel to the E field is given in units of Joules (J). This work appears as ELECTRICAL POTENTIAL ENERGY. 𝑊 = 𝑞 𝑉! − 𝑉! = 𝑞𝛥𝑉 This formula is NOT provided in the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. 9|PREPINEER.COM
psst… don’t forget to take notesÎ
It is important to remember that bringing a positive charge near another positive charge requires INPUT therefore, the work is POSITIVE. Bringing a negative charge, near a positive charge, RELEASES energy therefore, the work is NEGATIVE. The work W performed on a moving charge, 𝑄! a certain distance in a field created by charge, 𝑄! is represented with the electrostatic field formula in the form of Coulomb’s Law.
𝑊=
𝑄! 𝑄! 1 1 − 4𝜋𝜀 𝑟! 𝑟!
𝑊=
𝑘𝑄! 𝑄! 𝑟
where • 𝑄! = the charge at point A in Coulombs, C • 𝑄! = the charge at point B in Coulombs, C • 𝜀 = 8.85×10!!" F/m • 𝑘 = 9.0×10! Nm! /C ! = Coulomb’s Constant 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.
10|PREPINEER.COM
psst… don’t forget to take notesÎ
CONCEPT EXAMPLE: How much work must be done to bring a 4.0 𝜇C charged object to within 1.0 m of a 6.0 𝜇C charged object from a long way away? Is the work positive or negative?
SOLUTION: In this problem we will solve for the amount of work needed to move a charge based on electrostatics. In the problem, we are given: • A charge, 𝑄! = 4.0×10!! C • A charge, 𝑄! = 6.0×10!! C • The distance between the charges, 𝑟 = 1.0 m
Plugging these values into the formula for work and solving for W
𝑊=
𝑘𝑄! 𝑄! 𝑟
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psst… don’t forget to take notesÎ
(9.0×10! Nm! /C ! )(4.0×10!! C)(6.0×10!! C) 𝑊= 1.0 m 𝑊 = 0.216 J
Work is POSITIVE because we are bringing a positive charge near another positive charge requiring input, therefore the work is positive. 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.
12|PREPINEER.COM
psst… don’t forget to take notesÎ
