PPE04_Superposition - Electric Fields Electric Field of...

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© RHJansen Electric Fields Electric Fields Electric Field of Electric Field of Multiple Charged Points Multiple Charged Points (Superposition) (Superposition) © RHJansen
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Superposition Finding the electric field when there is more than one charge surrounding point P. Solve one charge at a time. Find the magnitude (value) and direction of each charge. This will give you the electric field vectors for every charge in the diagram. Then use vector addition to add all the electric field vectors to find the total electric field. © RHJansen
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Example 1 For the following charge configuration a. Determine the magnitude and direction of the electric field. © RHJansen I recommend that you draw the field vectors created by each charge before you start calculating. Visualize an imaginary test charge at point P . Ask yourself which direction the test charge will move due to each charge in the diagram. The positive charge (let’s call it q 1 ) will repel the test charge to the right. E 1 (caused by q 1 ) is to the right. q 1 E 1 q 2 E 2 The negative charge (let’s call it q 2 ) will attract the test charge to the right. E 2 (caused by q 2 ) is to the right. 0.2 μC P + 0.4 μC 2 m 2 m
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Example 1 For the following charge configuration a. Determine the magnitude and direction of the electric field. © RHJansen Now solve for E 1 and E 2 . q 1 E 1 q 2 E 2 0.2 μC P + 0.4 μC 2 m 2 m E 1 = κ θ 1 ρ 1 2 = 9 10 9 ( 29 0.4 10 -6 ( 29 2 ( 29 2 = 900 Ν Χ E 2 = k q 2 r 2 2 = 9 10 9 ( ) 0.2 10 - 6 ( ) 2 ( ) 2 = 450N C 900 450
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Example 1 For the following charge configuration a. Determine the magnitude and direction of the electric field. © RHJansen This is now a vector addition problem (see summer work) This vectors have the same direction, so they can be added very easily. E 1 E 2 P 900 450 We still need the direction of the electric field. This time we are lucky, there is no fancy trig involved. Just look at the arrows + x 900 + 450 = 1350 Ν Χ 900 450
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Example 1 For the following charge configuration b. A −0.3 μC charge is inserted at point P . Determine the magnitude and direction of the force on this charge. © RHJansen The electric field solved in part a (due to the original charges) pulls on the new charge P E = 1350 F E = θΕ = 0.3 10 -6 ( 29 1350 ( 29 = 4.05 10 -4 Ν The new charge is negative. Negative charges move opposite the field (they are the wrong way particle). The direction of force is x −0.3 μC F
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Example 2 Find the point where the electric field is zero. © RHJansen Before solving mathematically we need to know approximately where the point will be. There are three possible locations. First draw the vectors for the electric field, due to each charge, at all three locations. (Remember: you can place a positive test charge at all three points and ask which way it moves to determine field direction) The negative charge (black) will create a field that is toward itself at all three points. ( + test chg would move toward the negative charge)
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This note was uploaded on 07/21/2009 for the course PHYSICS 7B taught by Professor Packard during the Spring '08 term at University of California, Berkeley.

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PPE04_Superposition - Electric Fields Electric Field of...

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