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A21-Lect-4 - Chapter 26 The Electric Field Section 1,2,3...

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1/17/2009 1 Chapter 26 The Electric Field Section # 1,2,3 26.1 Electric field Models ¾ The electric field models can be divided into four types: 1. The electric field of a point charge. 2. The electric field of an infinitely long charged wire charged wire. 3. The electric field of an infinitely wide charged plane. 4. The electric field of a charged sphere. ¾ We’ve learned that the electric field is a vector We ve learned that the electric field is a vector. Therefore, the net electric field on q due to a group of point charges is the vector sum, = + + = + + = = i i q on q on q on net E E E q F q F q F E r r r .... .... 2 1 2 1 The principle of superposition
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1/17/2009 2 Limiting cases: we have two cases, some times we need to look at them, Very far from charged objects : In this case the object’s electric field represents an electric field of a point charge. Very close to charged objects : In this case the electric field depends on the object’s shape and on how the charge is distributed on that object. ¾ Suppose the source of an electric field is a group of point charges q 1 , q 2 , , then the electric field at any point in space (field point) is a 26.2 The Electric Field of Multiple Point Charges 1 , 2 ,…., then the electric field at any point in space (field point) is a superposition of the electric fields due to each source charge. ¾ The net electric field can be expressed in component form as: = + + = = + + = y i y y net x i x x net E E E E E E E E ) ( .... ) ( ) ( ) ( ) ( .... ) ( ) ( ) ( 2 1 y 2 1 x = + + = z i z z net E E E E ) ( .... ) ( ) ( ) ( 2 1 z represents the magnitude and direction
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1/17/2009 3 Ex: 3 equal point charges q are located on the y-axis at y=0 and y=± d . What is the electric field at a point on the x-axis ? Solve: From the fig. Are in the xy-plane. have equal magnitudes and angle θ Thus, by symmetry their y components cancel each other 3 2 1 , , E and E E r r r 3 1 E and E r r their y-components cancel each other. The only component we need to calculate is which is: x net E ) ( x x x x x x net E E E E E E ) ( 2 ) ( ) ( ) ( ) ( ) ( 1 2 3 2 1 + = + + = 2 0 2 2 2 0 2 2 4 1 4 1 ) ( x q r q E E x = = = π π 1 cos 1 cos ) ( q E E = = θ θ 2 / 3 2 2 0 2 / 1 2 2 2 2 0 1 2 / 1 2 2 1 2 2 1 2 1 0 1 1 ) ( 4 1 ) ( 4 1 ) ( ) ( cos 4 d x qx d x x d x q E d x x r x d x r r x x + = + + = + = = + = π π θ π
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1/17/2009 4 The Ex cont’d , Now, we can calculate x x x net E E E
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