eofcontdist

eofcontdist - . E i,x = k x r 2 sin , E i,y = k x r 2 cos...

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Phys 133 – General Physics III Finding the Electric Field of a Continuous Charge Distribution To set up an integral for a continuous charge distribution, follow this procedure: 1. Draw a diagram, choosing an appropriate coordinate system. Identify the point P at which the electric Feld is to be calculated. 2. Divide the total charge Q into small charge elements Δ Q . 3. Label all relevant distances on your diagram. M. Ouellette, 22 April 2007 Page 1 of 2
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4. Draw the electric feld vector v E i at point P For one small charge element Δ Q i . Indicate an angle θ on your diagram. 5. Write down an equation For the magnitude oF the electric feld at point P due to the small charge element Δ Q i . This is usually the equation For a point charge. E i = k Δ Q i r 2 6. Express Δ Q i in terms oF the charge density and a small distance element Δ x or Δ y and substitute this in your equation For the electric feld. λ = Q L Δ Q i = λ Δ x E i = Δ x r 2 7. Write down equations For the x - and y -components oF the electric feld using the angle
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Unformatted text preview: . E i,x = k x r 2 sin , E i,y = k x r 2 cos 8. Express the distance From P to Q i in terms oF the x- and y-coordinates, and substitute this in your equations For the electric feld components. r = r x 2 + y 2 E i,x = k x x 2 + y 2 sin , E i,y = k x x 2 + y 2 cos 9. Express sin and cos in terms oF the x- and y-coordinates and substitute these in your equations For the electric feld components. sin = x r x 2 + y 2 , cos = y r x 2 + y 2 E i,x = kx x ( x 2 + y 2 ) 3 / 2 , E i,y = ky x ( x 2 + y 2 ) 3 / 2 10. Convert your equations into integrals: change your small distance element x or y into a dierential d x or d y ; determine the limits oF integration. E x = k i L/ 2-L/ 2 x d x ( x 2 + y 2 ) 3 / 2 , E y = ky i L/ 2-L/ 2 d x ( x 2 + y 2 ) 3 / 2 M. Ouellette, 22 April 2007 Page 2 oF 2...
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eofcontdist - . E i,x = k x r 2 sin , E i,y = k x r 2 cos...

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