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Exercise:
A circular plate carries a uniform charge distribution with a total charge of Q. Derive a formal expression for the electric field at a po located along the central axis of the disc. Additionally discuss the case where the diameter of the plate is much greater than the distance from the plate.

Solution:
We divide the circular plate o concentric rings. center includegraphicswidth.mm#image_path:charged-plat# center The ring with radius r and width dr see figure produces an electric field dE at a po P with see field of charged ring dE k_C fracdQ dleftr^+d^right^/ Since the plate is uniformly charged the charge dQ can be expressed in terms of the total charge Q fracdQQ fracdAA fracpi r drpi R^ Longrightarrow dQ Q frac r drR^ where the expression dQ pi r dr is valid in the limit dr to . vspacemm The total electric field can be calculated by adding up the partial contributions of all rings. In the limit dr to this corresponds to the egral: E _^R dE _^R k_C fracdQ dleftr^+d^right^/ k_C d _^R fracQleftr^+d^right^/ frac r drR^ k_C frac Q dR^ _^R fracrleftr^+d^right^/ dr The antiderivative of the function fr fracrleftr^+d^right^/ is given by Fr -fracleftr^+d^right^/ as can be verified by taking the derivative of Fr: F'rfr. It follows for the electric field E k_C frac Q dR^ left -fracleftr^+d^right^/right_^R k_C frac Q dR^ left -fracleftR^+d^right^/+fracleftd^right^/right k_C frac Q dR^ left fracd-fracsqrtR^+d^right k_C frac QR^ left -fracsqrt+R/d^right With k_C /pi varepsilon_ this can be written as Ed fracQpi R^varepsilon_ left -fracsqrt+R/d^right fracvarepsilon_ fracQA left -fracsqrt+R/d^right fracsigmavarepsilon_ left -fracsqrt+R/d^right with the surface charge density sigmaQ/A. vspacemm If R gg d the second term in the parentheses ts to zero since R/d ts to infinity. The field can thus be approximated as E &approx fracsigmavarepsilon_ This means that the field of an infinetly large plate does not dep on the distance from the plate.
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Exercise:
A circular plate carries a uniform charge distribution with a total charge of Q. Derive a formal expression for the electric field at a po located along the central axis of the disc. Additionally discuss the case where the diameter of the plate is much greater than the distance from the plate.

Solution:
We divide the circular plate o concentric rings. center includegraphicswidth.mm#image_path:charged-plat# center The ring with radius r and width dr see figure produces an electric field dE at a po P with see field of charged ring dE k_C fracdQ dleftr^+d^right^/ Since the plate is uniformly charged the charge dQ can be expressed in terms of the total charge Q fracdQQ fracdAA fracpi r drpi R^ Longrightarrow dQ Q frac r drR^ where the expression dQ pi r dr is valid in the limit dr to . vspacemm The total electric field can be calculated by adding up the partial contributions of all rings. In the limit dr to this corresponds to the egral: E _^R dE _^R k_C fracdQ dleftr^+d^right^/ k_C d _^R fracQleftr^+d^right^/ frac r drR^ k_C frac Q dR^ _^R fracrleftr^+d^right^/ dr The antiderivative of the function fr fracrleftr^+d^right^/ is given by Fr -fracleftr^+d^right^/ as can be verified by taking the derivative of Fr: F'rfr. It follows for the electric field E k_C frac Q dR^ left -fracleftr^+d^right^/right_^R k_C frac Q dR^ left -fracleftR^+d^right^/+fracleftd^right^/right k_C frac Q dR^ left fracd-fracsqrtR^+d^right k_C frac QR^ left -fracsqrt+R/d^right With k_C /pi varepsilon_ this can be written as Ed fracQpi R^varepsilon_ left -fracsqrt+R/d^right fracvarepsilon_ fracQA left -fracsqrt+R/d^right fracsigmavarepsilon_ left -fracsqrt+R/d^right with the surface charge density sigmaQ/A. vspacemm If R gg d the second term in the parentheses ts to zero since R/d ts to infinity. The field can thus be approximated as E &approx fracsigmavarepsilon_ This means that the field of an infinetly large plate does not dep on the distance from the plate.
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Electrostatics
Tags
charge distribution, electric field
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Difficulty
(4, default)
Points
0 (default)
Language
ENG (English)
Type
Algebraic
Creator by
Decoration