EE204B_Lesson_06_ELectrostatic_Fields-1_

EE204B_Lesson_06_ELectrostatic_Fields-1_ - EE204B EE204B...

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Unformatted text preview: EE204B EE204B Electromagnetics Lesson 06 Electrostatic Fields Coulomb’s Law and Electric Field Intensity Instructor: Prof. Yong H. Won Lesson 06 Topics Electrostatic Fields (Chap. 4) • What is electrostatics? • Coulomb’s law • Electric field intensity EE204B Electromagnetics Prof. Yong H. Won Spring 2010 2 Lesson 06 What is electrostatics? EE204B Electromagnetics Prof. Yong H. Won Spring 2010 3 Lesson 06 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 4 Lesson 06 Coulomb’s Law Coulomb’s law quantifies the magnitude of the electrostatic force. Coulomb’s law gives the force (in Newtons) between charges q1 and q2, where r12 is the distance in meters between the charges, and k = 9 x 109 N·m2/C2. F12 = k q1q2 2 r12 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 5 Lesson 06 Force is a vector quantity. The equation on the previous slide gives the magnitude of the force If the charges are opposite in sign the force is magnitude of the force. If the charges are opposite in sign, the force is attractive; if the charges are the same in sign, the force is repulsive. Also, the constant k is equal to 1/4πε0, where ε0=8.85x10-12 C2/N·m2. One could write Coulomb’s Law like this… r q1q2 F12 = k 2 , r12 attractive for unlike To make this into a “really good” starting equation I should specify “repulsive for like,” but that makes it too wordy. You’ll just have to remember how to find the direction. Remember, a vector has a magnitude and a direction. di EE204B Electromagnetics Prof. Yong H. Won Spring 2010 6 Lesson 06 The equation is valid for point charges. If the charged objects are spherical and the charge is uniformly distributed and the charge is uniformly distributed, r12 is the distance between the the distance between the centers of the spheres. r12 + - If more than one charge is involved, the net force is the vector sum of all forces (superposition). For objects with complex shapes, you must add up all the forces acting on each separate charge (turns into calculus!). + + + - - - EE204B Electromagnetics Prof. Yong H. Won Spring 2010 7 Lesson 06 We could have agreed that in the formula for F, the symbols q1 and q2 stand for the magnitudes of the charges. In that case, the absolute value signs would be unnecessary. However, in later equations the sign of the charge will be important, so we really need to keep the magnitude part. On your homework diagrams, show both the magnitudes and signs of q1 and q2. Your starting equation sheet has this version of the equation: starting equation sheet has this version of the equation: q1q2 F12 = k 2 , r12 which gives you the magnitude F12 and tells you that you need to figure out the direction separately direction separately. EE204B Electromagnetics Prof. Yong H. Won Spring 2010 8 Lesson 06 Solving Problems Involving Coulomb’s Law and Vectors Example: Calculate the net electrostatic force on charge Q3 due to the charges Q1 and Q2. y Q3=+65μC 30 cm θ=30º Q2=+50μC 52 cm EE204B Electromagnetics Prof. Yong H. Won Q1=-86μC x Spring 2010 9 Lesson 06 Step 0: Think! This is a Coulomb’s Law problem (all we have to work with, so far). We only want the forces on Q3. Forces are additive, so we can calculate F23 and F13 and add the two. If we do our vector addition using components, we must resolve our forces we do our vector addition using components we must resolve our forces into their x- and y-components. EE204B Electromagnetics Prof. Yong H. Won Spring 2010 10 Lesson 06 Step 1: Diagram y Draw a representative sketch. Draw and label relevant quantities. Draw axes, showing origin and directions. F23 Q3=+65μC 30 cm F13 θ=30º Q2=+50μC 52 cm Q1=-86μC x Draw and label forces (only those on Q3). Draw components of forces which are not along axes. EE204B Electromagnetics Prof. Yong H. Won Spring 2010 11 Lesson 06 Step 2: Starting Equation y F23 Q3=+65μC 30 cm F13 θ=30º Q2=+50μC 52 cm Q1=-86μC x q1q2 F12 = k 2 r12 “Do I have to put in the absolute value signs?” EE204B Electromagnetics Prof. Yong H. Won Spring 2010 12 Lesson 06 Step 3: Replace Generic Quantities by Specifics y r Q2Q3 F23 = k 2 , r23 repulsive Q2Q3 =k 2 r23 F23 Q3=+65μC r23=30 cm F13 F23, y θ=30º Q1=-86μC 52 cm x Q2=+50μC F23, x = 0 (from diagram) F23,y = 330 N and F23,x = 0 N. EE204B Electromagnetics Prof. Yong H. Won Spring 2010 13 Lesson 06 Step 3 (continued) r Q1Q3 F13 = k 2 , r13 attractive F13, x Q1Q3 = + k 2 cos θ r13 y F23 Q3=+65μC r23=30 cm F13 θ=30º Q1=-86μC 52 cm (+ sign comes from diagram) Q2=+50μC x F13, y Q1Q3 = −k 2 sin θ r13 (- sign comes from diagram) comes from diagram) You would get F13,x = +120 N and F13,y = -70 N. EE204B Electromagnetics Prof. Yong H. Won Spring 2010 14 Lesson 06 Step 3: Complete the Math y F23 F3 F13 30 cm The net force is the vector sum of all the forces on Q3. Q3=+65μC θ=30º Q2=+50μC 52 cm F3x = F13,x + F23,x = 120 N + 0 N = 120 N F3y = F13,y + F23,y = -70 N + 330 N = 260 N Q1=-86μC x You know how to calculate the magnitude F3 and the angle between F3 and the x-axis. EE204B Electromagnetics Prof. Yong H. Won Spring 2010 15 Lesson 06 I did a sample Coulomb’s law calculation using three point charges. How do you apply Coulomb’s law to objects that contain distributions of do you apply Coulomb law to objects that contain distributions of charges? We’ll use another tool to do that… EE204B Electromagnetics Prof. Yong H. Won Spring 2010 16 Lesson 06 Vector Approach EE204B Electromagnetics Prof. Yong H. Won Spring 2010 17 Lesson 06 4-1 [Text Example 4.1] EE204B Electromagnetics Prof. Yong H. Won Spring 2010 18 Lesson 06 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 19 Lesson 06 4-2 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 20 Lesson 06 Electric Electric Field Intensity Coulomb's Law (demonstrated in 1785) shows that charged particles exert forces on each other over great distances. How does a charged particle "know" another one is “there?” EE204B Electromagnetics Prof. Yong H. Won Spring 2010 21 Lesson 06 Faraday, beginning in the 1830's, was the leader in developing the idea of the electric field. Here's the idea: F21 • A charged particle emanates a "field" into all space. • Another charged particle senses the field, and that the first one is there “knows” that the first one is there. + F31 + like charges repel repel F12 F13 unlike charges attract - EE204B Electromagnetics Prof. Yong H. Won Spring 2010 22 Lesson 06 Electric Field Intensity EE204B Electromagnetics Prof. Yong H. Won Spring 2010 23 Lesson 06 Electric Field Intensity EE204B Electromagnetics Prof. Yong H. Won Spring 2010 24 Lesson 06 The units of electric field are Newtons/Coulomb. r ⎡E ⎤ = ⎣⎦ r ⎡ F0 ⎤ ⎣⎦ [ q0 ] N = C Later you will learn that the units of electric field can also be expressed as volts/meter: [E] N V = = C m The electric field exists independent of whether there is a charged particle electric field exists independent of whether there is charged particle around to “feel” it. EE204B Electromagnetics Prof. Yong H. Won Spring 2010 25 Lesson 06 4-3 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 26 Lesson 06 4-4 [Text Practice 4.1] EE204B Electromagnetics Prof. Yong H. Won Spring 2010 27 Lesson 06 Remember: the electric field direction is the the electric field direction is the direction a + charge would feel a force. + A + charge would be repelled by another + charge. Therefore the direction of the electric field is away from positive (and towards negative). EE204B Electromagnetics Prof. Yong H. Won Spring 2010 28 Lesson 06 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 29 Lesson 06 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 30 Lesson 06 EE204B Electromagnetics Prof. Yong H. Won Spring 2010 31 ...
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