Question

A thin spherical shell with radius R1 = 2.00 cm is concentric with a larger thin spherical shell with radius R2 = 4.00 cm. Both shells are made of insulating material. The smaller shell has charge q1=+6.00nC distributed uniformly over its surface, and the larger shell has charge q2=−9.00nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. What is the electric potential due to the two shells at the following distance from their common center:
a. r = 0
b. r = 4.00cm
c. r = 6.00cm

Answer 1

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

$V=k( \frac{q1}{R1}+\frac{q2}{R2} )$

$V=(9x10^{9}[\frac{6x10^{-9} }{2x10^{-2} \\} + \frac{-9x10^{-9} }{4x10^{-2} } ]$

$V= 9x10^{9} * 7.5x10^{-8}$

V=675V

(b) r = 4.00cm. So, This point is outside of shell 1 but inside shell 2.

$V= k(\frac{q1}{r} + \frac{q2}{R2} )$

$V=(9x10^{9}) [\frac{6x10^{-9} }{4x10^{-2} } +\frac{-9x10^{-9} }{4x10^{-2} } ]$

$V= 9x10^{9} * -7.5x10^{-8}$

V=-675V

(c) r = 6.00cm. So, This point out of both shells.

$V = k(q1/r + q2/r) = \frac{k}{r} (q1 + q2)$

$V = \frac{9x10^{-9} }{6x10^{-2} } [6x10^{-9} + (-9x10^{-9}) ]\\V = -450 V$

Answer 2

Answer:.......

V = 675

V =675

V =450

Explanation:

Using the formula; V = z*sum(qi / ri) , where z is a constant.

q1 = +6.00nC ,$q1= 6*10^{-9}$. $R1 = 2.00cm ; R1 = 2*10^{-2}$.

q2= -9.00nC , $q2 = 9*10^{-9}$.  $R2 = 4.00cm ; R2 = 4*10^{-2}$.

when r =0, therefore radius is within both shells.

$V = z*(\frac{q1}{R1} + \frac{q2}{R2} )\\\\V = 9*10^{-9} *(\frac{6*10^{-9} }{2*10^{-2} } + \frac{-9*10^{-9} }{4*10^{-2} } )\\\\\V = 9*10^{9} * 7.5 *10^{-8}\\V = 675v$

when r = 4.00cm , therefore one R1 is outside the shell, while R2 is still within the shell

$V = z*(\frac{q1}{r} + \frac{q2}{R2} )\\\\V = 9*10^{-9} *(\frac{6*10^{-9} }{2*10^{-2} } + \frac{-9*10^{-9} }{4*10^{-2} } )\\\\\V = 9*10^{9} * 7.5 *10^{-8}\\V = 675v$

When r=6.00cm the 2 points are exceeding the shell

$V = z*(\frac{q1}{r} + \frac{q2}{r} )\\\\V = \frac{9*10^{-9} }{6*10^{-2} } ({6*10^{-9} }+{-9*10^{-9} } )\\\\\V = 450v$