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Exercise:
Derive a formal expression for the electric field produced by a uniformly charged ring on the ring's central axis.

Solution:
This is basically the same exercise as Charged ring and po charge see link where the po charge acts as the test charge to find the electric field vector. vspacemm We divide the ring o a large number N segments of equal length with charge dQsscQring/N. The force dF of one such segment on a test charge q is shown in the figure. center includegraphicswidth.mm#image_path:charged-ring-solution-# center The same amount of charge dQ'dQ on the opposite side of the ring exerts a force dF' on the po charge. The combined force sscdFnet is given by fracsscdFnet/dF fracdr Longrightarrow sscdFnet dF fracdr k_C fracdQ qr^ fracdr k_C fracsscQring q dN r^ k_C fracsscQring q dN leftR^ + d^right^/ There are N/ such pairs of opposite segments i.e. the total force on q is sscFnet fracN sscdFnet fracN k_C fracsscQring q dN leftR^ + d^right^/ k_C fracsscQring q dleftR^ + d^right^/ The electric field at the position of the test charge q is thus E fracsscFnetq k_C fracQ dleftR^+d^right^/ The direction of the field vector is along the central axis poing away from/towards the ring for a positively/negatively charged ring.
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Exercise:
Derive a formal expression for the electric field produced by a uniformly charged ring on the ring's central axis.

Solution:
This is basically the same exercise as Charged ring and po charge see link where the po charge acts as the test charge to find the electric field vector. vspacemm We divide the ring o a large number N segments of equal length with charge dQsscQring/N. The force dF of one such segment on a test charge q is shown in the figure. center includegraphicswidth.mm#image_path:charged-ring-solution-# center The same amount of charge dQ'dQ on the opposite side of the ring exerts a force dF' on the po charge. The combined force sscdFnet is given by fracsscdFnet/dF fracdr Longrightarrow sscdFnet dF fracdr k_C fracdQ qr^ fracdr k_C fracsscQring q dN r^ k_C fracsscQring q dN leftR^ + d^right^/ There are N/ such pairs of opposite segments i.e. the total force on q is sscFnet fracN sscdFnet fracN k_C fracsscQring q dN leftR^ + d^right^/ k_C fracsscQring q dleftR^ + d^right^/ The electric field at the position of the test charge q is thus E fracsscFnetq k_C fracQ dleftR^+d^right^/ The direction of the field vector is along the central axis poing away from/towards the ring for a positively/negatively charged ring.
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Attributes & Decorations
Branches
Electrostatics
Tags
charge distribution, coulomb, electric field
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Difficulty
(4, default)
Points
0 (default)
Language
ENG (English)
Type
Algebraic
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