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Chapter 16

Electric Forces and Fields

Table of ContentsSection 1 Electric Charge

Section 2 Electric Force

Section 3 The Electric Field

Section 1 Electric Charge

Chapter 16

Objectives- Understandthe basic properties of electric charge.
- Differentiatebetween conductors and insulators.
- Distinguishbetween charging by contact, charging by induction, and charging by polarization.

Section 1 Electric Charge

Chapter 16

Properties of Electric Charge- There are two kinds of electric charge.
- like charges repel
- unlike charges attract

- Electric charge is conserved.
- Positively charged particles are calledprotons.
- Uncharged particles are calledneutrons.
- Negatively charged particles are calledelectrons.

Section 1 Electric Charge

Chapter 16

Properties of Electric Charge, continued- Electric charge is quantized.That is, when an object is charged, its charge is always a multiple of afundamental unit of charge.
- Charge is measured in coulombs (C).
- The fundamental unit of charge, e, is the magnitude of the charge of a single electron or proton.
e = 1.602 176 x 10–19 C

Section 1 Electric Charge

Chapter 16

Transfer of Electric Charge- Anelectrical conductoris a material in which charges can move freely.
- Anelectrical insulatoris a material in which charges cannot move freely.

Section 1 Electric Charge

Chapter 16

Transfer of Electric Charge, continued- Insulators and conductors can be charged by contact.
- Conductors can be charged byinduction.
- Induction is a process of charging a conductor by bringing it near another charged object and grounding the conductor.

Section 1 Electric Charge

Chapter 16

Transfer of Electric Charge, continued- A surface charge can be induced on insulators bypolarization.
- With polarization, the charges within individual molecules are realigned such that the molecule has a slight charge separation.

Section 2 Electric Force

Chapter 16

Objectives- Calculateelectric force using Coulomb’s law.
- Compareelectric force with gravitational force.
- Applythe superposition principle to find the resultant force on a charge and to find the position at which the net force on a charge is zero.

Section 2 Electric Force

Chapter 16

Coulomb’s Law- Two charges near one another exert a force on one another called theelectric force.
- Coulomb’s law states that the electric force is propor-tional to the magnitude of each charge and inversely proportional to the square of the distance between them.

Section 2 Electric Force

Chapter 16

Coulomb’s Law, continued- The resultant force on a charge is the vector sum of the individual forces on that charge.
- Adding forces this way is an example of the principle of superposition.
- When a body is in equilibrium, the net external force acting on that body is zero.

Section 2 Electric Force

Chapter 16

Sample ProblemThe Superposition Principle

Consider three point charges at the corners of a triangle, as shown at right, where q1 = 6.00 10–9 C, q2 = –2.00 10–9 C, and q3 = 5.00 10–9 C. Find the magnitude and direction of the resultant force on q3.

Section 2 Electric Force

Chapter 16

Sample Problem, continuedThe Superposition Principle

1.Define the problem, and identify the known variables.

Given:

q1= +6.00 10–9 C r2,1 = 3.00 m

q2 = –2.00 10–9 C r3,2 = 4.00 m

q3 = +5.00 10–9 C r3,1= 5.00 m

q = 37.0º

Unknown:F3,tot= ? Diagram:

Section 2 Electric Force

Chapter 16

Sample Problem, continuedThe Superposition Principle

Tip:According to the superposition principle, the resultant force on the charge q3 is the vector sum of the forces exerted by q1 and q2 on q3. First, find the force exerted on q3 by each, and then add these two forces together vectorially to get the resultant force on q3.

2. Determine the direction of the forces by analyzing the charges.

The force F3,1 is repulsive because q1 and q3 have the same sign.

The force F3,2 is attractive because q2 and q3 have opposite signs.

Section 2 Electric Force

Chapter 16

Sample Problem, continuedThe Superposition Principle

3. Calculate the magnitudes of the forces with Coulomb’s law.

Section 2 Electric Force

Chapter 16

Sample Problem, continuedThe Superposition Principle

4. Find the x and y components of each force.

At this point, the direction each component must be taken into account.

F3,1:Fx = (F3,1)(cos 37.0º) = (1.08 10–8 N)(cos 37.0º)

Fx= 8.63 10–9 N

Fy = (F3,1)(sin 37.0º) = (1.08 10–8 N)(sin 37.0º)

Fy = 6.50 10–9 N

F3,2: Fx= –F3,2 = –5.62 10–9 N

Fy = 0 N

Section 2 Electric Force

Chapter 16

Sample Problem, continuedThe Superposition Principle

5. Calculate the magnitude of the total force acting in both directions.

Fx,tot = 8.63 10–9 N – 5.62 10–9 N = 3.01 10–9 N

Fy,tot = 6.50 10–9 N + 0 N = 6.50 10–9 N

Section 2 Electric Force

Chapter 16

Sample Problem, continuedThe Superposition Principle

6. Use the Pythagorean theorem to find the magni-tude of the resultant force.

Section 2 Electric Force

Chapter 16

Sample Problem, continuedThe Superposition Principle

7. Use a suitable trigonometric function to find the direction of the resultant force.

In this case, you can use the inverse tangent function:

Section 2 Electric Force

Chapter 16

Coulomb’s Law, continued- The Coulomb force is a field force.
- A field force is a force that is exerted by one object on another even though there is no physical contact between the two objects.

Section 3 The Electric Field

Chapter 16

Objectives- Calculateelectric field strength.
- Draw and interpretelectric field lines.
- Identifythe four properties associated with a conductor in electrostatic equilibrium.

Section 3 The Electric Field

Chapter 16

Electric Field Strength- Anelectric fieldis a region where an electric force on a test charge can be detected.
- The SI units of the electric field, E, are newtons per coulomb (N/C).
- The direction of the electric field vector, E,is in the direction of the electric force that would be exerted on a small positive test charge.

Section 3 The Electric Field

Chapter 16

Electric Field Strength, continued- Electric field strength depends on charge and distance. An electric field exists in the region around a charged object.
- Electric Field Strength Due to a Point Charge

Section 3 The Electric Field

Chapter 16

Sample ProblemElectric Field Strength

A charge q1 = +7.00 µC is at the origin, and a charge q2 = –5.00 µC is on the x-axis 0.300 m from the origin, as shown at right. Find the electric field strength at point P,which is on the y-axis 0.400 m from the origin.

Section 3 The Electric Field

Chapter 16

Sample Problem, continuedElectric Field Strength

1.Define the problem,andidentifytheknownvariables.

Given:

q1= +7.00 µC = 7.00 10–6 C r1 = 0.400 m

q2 = –5.00 µC = –5.00 10–6 C r2 = 0.500 m

- = 53.1º
Unknown:

E at P (y = 0.400 m)

Tip:Apply the principle of superposition. You must first calculate the electric field produced by each charge individually at point P and then add these fields together as vectors.

Section 3 The Electric Field

Chapter 16

Sample Problem, continuedElectric Field Strength

2. Calculate the electric field strength produced by each charge. Because we are finding the magnitude of the electric field, we can neglect the sign of each charge.

Section 3 The Electric Field

Chapter 16

Sample Problem, continuedElectric Field Strength

3. Analyze the signs of the charges.

The field vector E1 at P due to q1 is directed vertically upward, as shown in the figure, because q1 is positive. Likewise, the field vector E2at P due to q2 is directed toward q2 because q2 is negative.

Section 3 The Electric Field

Chapter 16

Sample Problem, continuedElectric Field Strength

4. Find the x and y components of each electric field vector.

For E1:Ex,1 = 0 N/C

Ey,1 = 3.93 105 N/C

For E2:Ex,2= (1.80 105 N/C)(cos 53.1º) = 1.08 105 N/C

Ey,1= (1.80 105 N/C)(sin 53.1º)= –1.44 105 N/C

Section 3 The Electric Field

Chapter 16

Sample Problem, continuedElectric Field Strength

5. Calculate the total electric field strength in both directions.

Ex,tot = Ex,1 + Ex,2= 0 N/C + 1.08 105 N/C

= 1.08 105 N/C

Ey,tot = Ey,1 + Ey,2= 3.93 105 N/C – 1.44 105 N/C

= 2.49 105 N/C

Section 3 The Electric Field

Chapter 16

Sample Problem, continuedElectric Field Strength

6. Use the Pythagorean theorem to find the magnitude of the resultant electric field strength vector.

Section 3 The Electric Field

Chapter 16

Sample Problem, continuedElectric Field Strength

7. Use a suitable trigonometric function to find the direction of the resultant electric field strength vector.

In this case, you can use the inverse tangent function:

Section 3 The Electric Field

Chapter 16

Sample Problem, continuedElectric Field Strength

8. Evaluate your answer.

The electric field at point P is pointing away from the charge q1, as expected, because q1 is a positive charge and is larger than the negative charge q2.

Section 3 The Electric Field

Chapter 16

Electric Field Lines- The number of electric field lines is proportional to the electric field strength.
- Electric field lines are tangent to the electric field vector at any point.

Section 3 The Electric Field

Chapter 16

Conductors in Electrostatic Equilibrium- The electric field is zero everywhere inside the conductor.
- Any excess charge on an isolated conductor resides entirely on the conductor’s outer surface.
- The electric field just outside a charged conductor is perpendicular to the conductor’s surface.
- On an irregularly shaped conductor, charge tends to accumulate where the radius of curvature of the surface is smallest, that is, at sharp points.

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