Asked by AAA
The total charge on a uniformly charged ring with diameter 26.0 cm is −57.3 µC. What is the magnitude of the electric field along the ring's axis at the following distances from its center?
(a) 2.60 cmN/C, (b) 13.0 cmN/C, (c) 26.0 cm N/C, (d) 2.08 m N/C
(a) 2.60 cmN/C, (b) 13.0 cmN/C, (c) 26.0 cm N/C, (d) 2.08 m N/C
Answers
Answered by
Damon
a = radius = 0.13 meter
x = distance from center along that center line
is charge in coulombs
I do not know if you have been given the formula, it is:
Ex = (1/ [4pi eo] ) Q x / (x^2+a^2)^1.5
================================
If you need the derivation
dQ at spot along ring
let k = (1/ [4pi eo] ) for now
dE = k dQ /(x^2+a^2)
x component of that (the y components cancel of course)
dEx =dE cos angle = k dQ /(x^2+a^2) * x/sqrt(x^2+a^2)
= k dQ [ x/ (x^2+a^2)^1.5 ]
integrate dQ at our chosen constant x
Ex = k Q [ x/ (x^2+a^2)^1.5 ]
x = distance from center along that center line
is charge in coulombs
I do not know if you have been given the formula, it is:
Ex = (1/ [4pi eo] ) Q x / (x^2+a^2)^1.5
================================
If you need the derivation
dQ at spot along ring
let k = (1/ [4pi eo] ) for now
dE = k dQ /(x^2+a^2)
x component of that (the y components cancel of course)
dEx =dE cos angle = k dQ /(x^2+a^2) * x/sqrt(x^2+a^2)
= k dQ [ x/ (x^2+a^2)^1.5 ]
integrate dQ at our chosen constant x
Ex = k Q [ x/ (x^2+a^2)^1.5 ]
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