Electric Charges and Fields: Lecture Overview Chapter 19

Lecture 3: Chapter 19 Cont Electric Charges, Forces and Electric Fields

Agenda • Compare the electric force to Gravitational force • Superposition of forces • Electric Field • Superposition of Electric Field • Shielding and Charging by Induction • Electric Flux and Gauss’s Law

What did we study last lesson? • Electric Charges: positive and neg. Electron (neg) and proton (pos) • Charges are conserved • Polarization • Insulators & Conductors • Coulomb’s Law

Recap • Two rods and a cat fur: • Glass Rod (positive charges) • Rubber Rod (negative)

Charging by Conduction

Recap • Coulomb’s Law • (electrostatic force between point charges)

Recap • Polarization is realignment of charge within individual molecules. • Produces induced charge on the surface of insulators. • how e.g. rubber or glass can be used to supply electrons.

? + Student Discussion A positively charged object hanging from a string is brought near a non conducting object (ball). The ball is seen to be attracted to the object. • Explain why it is not possible to determine whether the object is negatively charged or neutral. • What additional experiment is needed to reveal the electrical charge state of the object?

- + Cont • Two possibilities: • Attraction between objects of unlike charges. Attraction between a charged object and a neutral object subject to polarization. + - + - + - + + -

? 0 +- + + + + - + -+- What additional experiment is needed to reveal the electrical charge state of the object? • Two Experiments: • Bring known neutral ball near the object and observe whether there is an attraction. Bring a known negatively charge object near the first one. If there is an attraction, the object is neutral, and the attraction is achieved by polarization. -

Question • Name the first action at a distance force you have encountered in physics so far. Electric Force

Electrical Forces are Field Forces • This is the second example of a field force • Gravity was the first • Remember, with a field force, the force is exerted by one object on another object even though there is no physical contact between them • There are some important similarities and differences between electrical and gravitational forces

Electrical Force Compared to Gravitational Force • Both are inverse square laws • The mathematical form of both laws is the same • Masses replaced by charges • G replaced by ke • Electrical forces can be either attractive or repulsive • Gravitational forces are always attractive

Consider a proton (mp=1.67x10-27 kg; qp=+1.60x10-19 C) and an electron (me=9.11x10-31 kg; qe=-1.60x10-19 C) separated by 5.29x10-11 m. The particles are attracted to each other by both the force of gravity and by Coulomb’s law force. Which of these has the larger magnitude? • Gravitational force • B. Coulomb’s law force Example: Student Participation

Which of these has the larger magnitude? • Gravitational force • B. Coulomb’s law force Pg. 659. By what factor the electric force is greater than the gravitational force? We will mostly ignore gravitational effects when we consider electrostatics.

Problem solving steps • Visualize problem – labeling variables • Determine which basic physical principle(s) apply • Write down the appropriate equations using the variables defined in step 1. • Check whether you have the correct amount of information to solve the problem (same number of knowns and unknowns). • Solve the equations. • Check whether your answer makes sense (units, order of magnitude, etc.).

Superposition Principle • From observations: one finds that whenever multiple charges are present, the net force on a given charge is the vector sum of all forces exerted by other charges. • Electric force obeys a superposition principle.

The Superposition Principle • How to work the problem? • Find the electrical forces between pairs of charges separately • Then add the vectors • Remember to add the forces as vectors

Superposition Principle Example: 3 charges in a line F13 F12 q1 = 6.00 uC q2 = 1.50 uC q3 = -2.00 uC d1 = 3.00 cm d2 = 2.00 cm K = 8.99 x 109 N.m2 /C2

Example: Superposition Principle (Not along a line) Consider three point charges at the corners of a triangle, as shown in the next slide. Find the resultant force on q3 if q1 = 6.00 x 10-9 C q2 = -2.00 x 10-9 C q3 = 5.00 x 10-9 C

Superposition Principle Example • The force exerted by q1 on q3 is • The force exerted by q2 on q3 is • The total force exerted on q3 is the vector sum of and

Consider three point charges at the corners of a triangle, as shown below. Find the resultant force on q3. Solution:

Spherical Charge Distribution • Students need to read and work on pg 663/664

Electric Field - Discovery • Electric forces act through space even in the absence of physical contact. • Suggests the notion of electrical field (first introduced by Michael Faraday (1791-1867).

Electric Field • An electric field is said to exist in a region of space surrounding a charged object. • If another charged object enters a region where an electrical field is present, it will be subject to an electrical force.

Electric Field • A charged particle, with charge Q, produces an electric field in the region of space around it • A small test charge, qo, placed in the field, will experience a force. • “The electric field is the force per charge at a given location”

Electric Field • Mathematically, • SI units are N / C Given: One finds:

Electric Field • Use this for the magnitude of the field • The electric field is a vector quantity • The direction of the field is defined to be the direction of the electric force that would be exerted on a small positive test charge placed at that point

Direction of Electric Field • The electric field produced by a negative charge is directed toward the charge • A positive test charge would be attracted to the negative source charge

Direction of Electric Field, cont • The electric field produced by a positive charge is directed away from the charge • A positive test charge would be repelled from the positive source charge

More About a Test Charge and The Electric Field • The test charge is required to be a small charge • It can cause no rearrangement of the charges on the source charge • The electric field exists whether or not there is a test charge present • The Superposition Principle can be applied to the electric field if a group of charges is present

Electric Fields and Superposition Principle • The superposition principle holds when calculating the electric field due to a group of charges • Find the fields due to the individual charges • Add them as vectors • Use symmetry whenever possible to simplify the problem

Electric Field, final

Problem Solving Strategy • Draw a diagram of the charges in the problem • Identify the charge of interest • You may want to circle it • Units – Convert all units to SI • Need to be consistent with ke

Problem Solving Strategy, cont • Apply Coulomb’s Law • For each charge, find the force on the charge of interest • Determine the direction of the force • Sum all the x- and y- components • This gives the x- and y-components of the resultant force • Find the resultant force by usingthe Pythagorean theorem and trig

Problem Solving Strategy, Electric Fields • Calculate Electric Fields of point charges • Use the equation to find the electric field due to the individual charges • The direction is given by the direction of the force on a positive test charge • The Superposition Principle can be applied if more than one charge is present

Examples of Electric Field • E will be constant (a charged plane) • E will decrease with distance as (a point charge) • E in a line charge, decrease with a distance 1/r

Example: • An electron moving horizontally passes between two horizontal planes, the upper plane charged negatively, and the lower positively. - - - - - - - - - - - - - - - - - - - - - - vo - + + + + + + + + + + + + + + + + + + + + + +

Example: • A uniform, upward-directed electric field exists in this region. This field exerts a force on the electron. Describe the motion of the electron in this region. - - - - - - - - - - - - - - - - - - - - - - vo - + + + + + + + + + + + + + + + + + + + + + +

- - - - - - - - - - - - - - - - - - - - - - vo - + + + + + + + + + + + + + + + + + + + + + + Observations: • Horizontally: • No electric field • No force • No acceleration • Constant horizontal velocity

- - - - - - - - - - - - - - - - - - - - - - vo - + + + + + + + + + + + + + + + + + + + + + + Observations: • Vertically: • Constant electric field • Constant force • Constant acceleration • Vertical velocity increase linearly with time.

- - - - - - - - - - - - - - - - - - - - - - - + + + + + + + + + + + + + + + + + + + + + + Conclusions: The charge will follow a parabolic path downward. Motion similar to motion under gravitational field only except the downward acceleration is now larger.

y E1 E P E2 0.400 m x q1 0.300 m q2 Example: Electric Field Due to Two Point Charges (Superposition of Fields) Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m.

Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m. Observations: • First find the Electric field at point P due to charge q1 and q2. • Field E1 at P due to q1 is vertically upward. • Field E2 at due to q2 is directed towards q2. • The net field at point P is the vector sum of E1 and E2. • The magnitude is obtained with

Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m. Set up the problem: q1=7.00 mC q2=-10.00 mC K = 8.99 x 109 N. m2 /C2 r1 = 0.400m r2= ? What do we do?

Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m. • Calculate r2 • Calculate the electric field at P • I will work set it up, but you will need to finish (it looks like example in pg. 669)

Question: Charge q1=7.00 mC is at the origin, and charge q2=-10.00 mC is on the x axis, 0.300 m from the origin. Find the electric field at point P, which has coordinates (0,0.400) m. Solution:

Electric Field Lines • A convenient aid for visualizing electric field patterns is to draw lines pointing in the direction of the field vector at any point • These are called electric field lines and were introduced by Michael Faraday

Electric Field Lines, cont. • The field lines are related to the field in the following manners: • The electric field vector, , is tangent to the electric field lines at each point • The number of lines per unit area through a surface perpendicular to the lines is proportional to the strength of the electric field in a given region

Electric Field Line Patterns • Point charge • The lines radiate equally in all directions • For a positive source charge, the lines will radiate outward

Electric Charges and Fields: Lecture Overview Chapter 19