Chapter 22b - Last class we found that a uniformly charged...

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dx Last class we found that a uniformly charged thin rod of total charge q and length L , lying along the x axis (as shown), ˆ E(x)i produces an electric field at points on the x axis (not on the rod) of magnitude, () q E(x) k xx L = + q x L To arrive at this result we had to divide the rod into infinitesimal charge elements dx , each of which contains an infinitesimal quantity of charge dq . x dq
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Where r is the distance between dq and q . p 2 (q)(dq) dF k r = The infinitesimal electric field magnitude at the position (x p ) of q (our unit positive test charge) is then by definition this force divided by the unit charge, p p 2 dF dq dE(x ) k qr == x x p dx Each such infinitesimal charge element , dq, exerts an infinitesimal force , dF p , on our (fictitious) unit, positive, test charge q located at position x p on the axis. The magnitude of that force is given by Coulomb’s law, dq q r
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We need to sum the electric field components at x p due to each charge element dq along the rod but each dq element is at a different distance from q . x x p dx dq q r We can write their distance to x p as r = x p –x . Then, x p 2 p dq dE(x ) k (x x) = To sum the infinitesimal electric field components at x p getting the net electric field at x p we must integrate the infinitesimal charge elements over the length of the rod .
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E 2 p 0r o d dq dE k (x x) = ∫∫ But to proceed our variable of integration must be connected to what changes in performing the sum (i.e. the position of dq ). What makes this connection is the linear charge density λ since, qd q dq dx Ld x λ= = Making this substitution for dq above, p 22 pp rod rod dx dx E(x ) k k λ == λ −−
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Previously we defined our coordinate system so that one end of the rod lay at the origin but this need not be the case. Let’s now instead consider the center of the rod to be at the origin. x x p dx dq q r x We still have that r = x p x (for dq elements to the left of the origin the position x is a negative number). So the only change from before is the limits on the integration, which now must run from – L/2 to + L/2 so, L 2 p 2 L p 2 dx E(x ) k (x x)
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L L 2 2 p 2 L L pp 2 2 dx 1 E(x ) k k (x x) x x −− The integral is the same as last time with the same solution p 11 k LL xx 22 −+ p k +− ⎛⎞ ⎜⎟ ⎝⎠ p k +
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p pp L E(x ) k LL xx 22 ⎛⎞ −+ ⎜⎟ ⎝⎠ p 2 2 p L k L x 4 But so, q L λ= p qL k q k L 44 ⎡⎤ ⎢⎥ == Since the point x p is arbitrary we drop the subscript getting, 2 2 kq E(x) L x 4 =
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() q E(x) k xx L = x x 2 2 q k L x 4 = This result seems different from what we got last time.
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This note was uploaded on 03/12/2010 for the course PHY PHY taught by Professor Mueller during the Spring '09 term at University of Florida.

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Chapter 22b - Last class we found that a uniformly charged...

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