Answer:
a)v= 675V
b) v=-675V
c)v=-450V
Explanation:
Formula: V = k*sum(qi/ri)
Given:
q1 = +6.00nC --> R1 = 2.00cm
q2 = -9.00nC --> R2 = 4.00cm
(a) r = 0. So, inside both shells
[tex]V=k( \frac{q1}{R1}+\frac{q2}{R2} )[/tex]
[tex]V=(9x10^{9}[\frac{6x10^{-9} }{2x10^{-2} \\} + \frac{-9x10^{-9} }{4x10^{-2} } ][/tex]
[tex]V= 9x10^{9} * 7.5x10^{-8}[/tex]
V=675V
(b) r = 4.00cm. So, This point is outside of shell 1 but inside shell 2.
[tex]V= k(\frac{q1}{r} + \frac{q2}{R2} )[/tex]
[tex]V=(9x10^{9}) [\frac{6x10^{-9} }{4x10^{-2} } +\frac{-9x10^{-9} }{4x10^{-2} } ][/tex]
[tex]V= 9x10^{9} * -7.5x10^{-8}[/tex]
V=-675V
(c) r = 6.00cm. So, This point out of both shells.
[tex]V = k(q1/r + q2/r) = \frac{k}{r} (q1 + q2)[/tex]
[tex]V = \frac{9x10^{-9} }{6x10^{-2} } [6x10^{-9} + (-9x10^{-9}) ]\\V = -450 V[/tex]
Answer:.......
V = 675
V =675
V =450
Explanation:
Using the formula; V = z*sum(qi / ri) , where z is a constant.
q1 = +6.00nC ,[tex]q1= 6*10^{-9}[/tex]. [tex]R1 = 2.00cm ; R1 = 2*10^{-2}[/tex].
q2= -9.00nC , [tex]q2 = 9*10^{-9}[/tex]. [tex]R2 = 4.00cm ; R2 = 4*10^{-2}[/tex].
when r =0, therefore radius is within both shells.
[tex]V = z*(\frac{q1}{R1} + \frac{q2}{R2} )\\\\V = 9*10^{-9} *(\frac{6*10^{-9} }{2*10^{-2} } + \frac{-9*10^{-9} }{4*10^{-2} } )\\\\\V = 9*10^{9} * 7.5 *10^{-8}\\V = 675v[/tex]
when r = 4.00cm , therefore one R1 is outside the shell, while R2 is still within the shell
[tex]V = z*(\frac{q1}{r} + \frac{q2}{R2} )\\\\V = 9*10^{-9} *(\frac{6*10^{-9} }{2*10^{-2} } + \frac{-9*10^{-9} }{4*10^{-2} } )\\\\\V = 9*10^{9} * 7.5 *10^{-8}\\V = 675v[/tex]
When r=6.00cm the 2 points are exceeding the shell
[tex]V = z*(\frac{q1}{r} + \frac{q2}{r} )\\\\V = \frac{9*10^{-9} }{6*10^{-2} } ({6*10^{-9} }+{-9*10^{-9} } )\\\\\V = 450v[/tex]
