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Unformatted text preview: ,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 distributi...
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This note was uploaded on 05/14/2010 for the course EE EE204 taught by Professor Parkkyungsoo during the Spring '10 term at Korea University.
- Spring '10