Question

Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a

Answer 1

electric potential, V = -q(a²- b²)/8π∈₀r³

Step-by-step explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

consider the attached diagram below

the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below

Va = q/4π∈₀ [1/(a² + b²)¹/²]

`Va = (q)/(4\pi e0) * (1)/((a^(2) + r^(2) )^(1/2) )`

Also

the electric potential at point p, distance r from the center of the inner charged ring with radius b is

`Vb = (-q)/(4\pi e0) * (1)/((b^(2) + r^(2) )^(1/2) )`

Sum of the potential at point p is

V = Va + Vb

that is

`V = (q)/(4\pi e0) * (1)/((a^(2) + r^(2) )^(1/2) ) + (-q)/(4\pi e0 ) * (1)/((b^(2) + r^(2) )^(1/2) )`

`V = (q)/(4\pi e0) * (1)/((a^(2) + r^(2) )^(1/2) ) - (q)/(4\pi e0 ) * (1)/((b^(2) + r^(2) )^(1/2) )`

`V = (q)/(4\pi e0) * [(1)/((a^(2) + r^(2) )^(1/2) ) - (1)/((b^(2) + r^(2) )^(1/2) )]`

the expression below can be written as the equivalent

`(1)/((a^(2) + r^(2) )^(1/2) ) = (1)/((r^(2) + a^(2) )^(1/2) ) = \frac{1}{{r(1^(2) + (a^(2) )/(r^(2) ) )}^(1/2) }`

likewise,

`(1)/((b^(2) + r^(2) )^(1/2) ) = (1)/((r^(2) + b^(2) )^(1/2) ) = \frac{1}{{r(1^(2) + (b^(2) )/(r^(2) ) )}^(1/2) }`

hence,

`V = (q)/(4\pi e0) * [\frac{1}{{r(1^(2) + (a^(2) )/(r^(2) ) )}^(1/2) } - \frac{1}{{r(1^(2) + (b^(2) )/(r^(2) ) )}^(1/2) }]`

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

`V = (q)/(4\pi e0 r) * [\frac{1}{{(1^(2) + (a^(2) )/(r^(2) ) )}^(1/2) } - \frac{1}{{(1^(2) + (b^(2) )/(r^(2) ) )}^(1/2) }]`

by reciprocal rule

1/a² = a⁻²

`V = (q)/(4\pi e0 r) * [{(1^(2) + (a^(2) )/(r^(2) ) )}^(-1/2) - {(1^(2) + (b^(2) )/(r^(2) ) )}^(-1/2)]`

by binomial expansion of fractional powers

where
`(1+a)^(n) =1+na+(n(n-1)a^(2) )/(2!)+ (n(n-1)(n-2)a^(3))/(3!)+...`

if we expand the expression we have the equivalent as shown

`{(1^(2) + (a^(2) )/(r^(2) ) )}^(-1/2) = (1-(a^(2) )/(2r^(2) ) )`

also,

`{(1^(2) + (b^(2) )/(r^(2) ) )}^(-1/2) = (1-(b^(2) )/(2r^(2) ) )`

the above equation becomes

`V = (q)/(4\pi e0 r) * [((1-(a^(2) )/(2r^(2) ) ) - (1-(b^(2) )/(2r^(2) ) )]`

`V = (q)/(4\pi e0 r) * [1-(a^(2) )/(2r^(2) ) - 1+(b^(2) )/(2r^(2) )]`

`V = (q)/(4\pi e0 r) * [-(a^(2) )/(2r^(2) ) +(b^(2) )/(2r^(2) )]\\\\V = (q)/(4\pi e0 r) * [(b^(2) )/(2r^(2) ) -(a^(2) )/(2r^(2) )]`

`V = (q)/(4\pi e0 r) * (1)/(2r^(2) ) *(b^(2) -a^(2) )`

`V = (q)/(8\pi e0 r^(3) ) * (b^(2) -a^(2) )`

Answer

`V = (q (b^(2) -a^(2) ))/(8\pi e0 r^(3) )`

OR

`V = (-q (a^(2) -b^(2) ))/(8\pi e0 r^(3) )`