PHY2049ch22B%281-15-10%290

PHY2049ch22B%281-15-10%290 - Last class we asked what is...

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dx Last class we asked what is the electrostatic field along the axis of a uniformly charged thin rod of total charge q and length L . p x + q x L Since we know how to evaluate the electrostatic force between point particles we divide the rod into infinitesimal lengths dx that contain infinitesimal charge dq . x dq To evaluate this we placed the rod on the x axis with one end at the origin. To probe for the electric field we placed a unit, positive, test charge at point x p along the axis. o q p x o q
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Where r is the distance between dq and q o . o p 2 (q )(dq) dF k r = The infinitesimal electric field magnitude at the position x p is then by definition this force divided by the unit (test) charge, p p 2 o 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 o located at position x p on the axis. The magnitude of that force is given by Coulomb’s law, dq q o 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 get the net electric field magnitude at x p we must sum the contributions to this field from all the infinitesimal charge elements i.e. we must integrate this function 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 and using for the integration limits the region over which there is charge on the rod (i.e. 0 to L ) we have, LL p 22 pp 00 dx dx E(x ) k k λ == λ −−
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() L 2 0 p dx Ek xx L P pp 0 11 1 k x xL ⎡⎤ ⎢⎥ −− ⎣⎦ Now, 2 P p dx (x x) = Then 1 p p d1 d [(x x) ] dx (x dx =− ⎜⎟ since, 2 p 1( x x) ( 1) = 2 p 1 = So,
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Since the point x P was arbitrary, we drop the subscript giving the field magnitude everywhere along the x axis (but not on or in the rod). () 11 E(x) k xL x = λ− ˆ E(x)i ˆ The term in brackets simplifies: ( ) xx L L x x x −= = −− and since this reduces to, q L λ= q k xx L = With vector electric field direction away from the rod if q is positive (assumed above) and toward the rod if q is negative.
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Example 2: we evaluate the electric field along a line that is perpendicular to and bisects the rod, x z We again divide the rod into infinitesimal charge elements and calculate the field at z P due to each such element.
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PHY2049ch22B%281-15-10%290 - Last class we asked what is...

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