Chapter_2

# Chapter_2 - EP 106 General Physics II Chapter 2 Electric...

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EP 106 General Physics II Chapter 2: Electric Field Lecture Outline 1. The Definition of Electric Field 2. Electric Field Lines 3. The Electric Field Due to Point Charges 4. The Electric Field Due to Continuous Charge Distributions 5. The Force on Charges in Electric Fields Electric field A charge can experience an electrostatic force due to the presence of other charges. The idea of a field is required in order to explain the "action at a distance." Recall the idea of gravitational field. m g F g In this view, Earth creates a force on the mass m. This is “insane.” Earth isn’t even touching the mass. So we introduce the idea of a gravitational field. Now we take the view that the field due to Earth, g, is exerting the gravitational force on the mass. Since the force is , the gravitational field is defined as, G F g = m G g G g G F g m . Let’s take the same approach with the electric force. q 1 E F e q Instead of thinking of q exerting the force on q 1 , we think of q creating a field and the field exerting the force on q 1 . Mathematically, we can write Gaziantep University Faculty of Engineering Department of Engineering Physics 1 which is the definition of the electric field. The electric field is a vector, and its direction is the same as the direction of the force F G on a positive test charge. It has units Newton per coulomb (N/C) 1 1 e e F Fq EE q =⇒ = G GG G Note: -The surrounding charges (not q) is the ones that create an electric field. -Similarly between electric field and gravitational field

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EP 106 General Physics II The electric field due to a point charge Gaziantep University Faculty of Engineering Department of Engineering Physics 2 The electric field due to N charged particles Definition of the electric field r r kq E ˆ 2 = K = = n i i i i r r kq E 1 0 2 0 ˆ K Example: A point charge Q 1 = + 10.0 C is at the origin, and a second charge Q 2 = -5.0 C is placed on the y-axis at y = 1.0 m. What is the total electric field at the point P with coordinates x = 2.0 m, y = 1.0 m ? We use Coulomb laws to find the field due to each charge, and then we find the vector sum. First, we found the field 1 E K due to Q 1 C N j i j i m m C C m N r r Q k E / 10 ] ˆ 8 . 0 ˆ 6 . 1 [ 5 ) ˆ ˆ 2 ( ] ) 0 . 1 ( ) 0 . 2 [( ) 0 . 10 ( ) . 10 0 . 9 ( ˆ 10 2 2 2 2 9 1 2 1 1 1 × + = + + × = = K The field at P due to the negative charge Q 2 is in the –x direction: ) / 10 1 . 1 ( ˆ ˆ ) 0 . 2 ( ) 0 . 5 ( ) . 10 0 . 9 ( ˆ 10 2 2 2 9 2 2 2 2 2 C N i i m C C m N r r Q k E × = × = = K C N j i C N i j i E E E / 10 ] ˆ 8 . 0 ˆ 5 . 0 [ / 10 ] ˆ 1 . 1 ˆ 8 . 0 ˆ 6 . 1 [ 10 10 2 1 × + = × + = + = K K K or E K =(0.94x10 10 )N/C, in direction j i ˆ 8 . 0 ˆ 5 . 0 + Electric field lines We need a more descriptive image of the field. The most useful idea is “Electric Field Lines.” Field Lines or Lines of Force are used to visualize the field. The rules for drawing them are: 1. The tangent to the field line points in the direction of the force on a positive test charge.
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Chapter_2 - EP 106 General Physics II Chapter 2 Electric...

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