Question

(a) A small plastic bead with a charge of -60.0 nC is at the center of an insulating rubber spherical shell with an inner radius of 20.0 cm and an outer radius of 39.0 cm. The rubber material of the spherical shell is charged, with a uniform volume charge m/s)? density of -1.80 HC/m3. A proton moves in a circular orbit just outside the spherical shell. What is the speed of the proton (in (b) What If? Suppose the spherical shell carries a positive charge density instead. What is the maximum value the charge density (in HC/m3) the spherical shell can have below which a proton can orbit the spherical shell? HC/m

Answer 1

(a)

given

q = -60 nC

a = 20 cm

b = 39 cm

= -1.80 uC/m³

Given q = -60 nC a = 20 cm b = 39 cm rho = -1.80 uC/m^3 total charge on the shell is Q = rho*volume = rho*4/3*phi*(b^3 - a^3) Q = -1.80*10^-6 *4/3*3.14*((0.39)^3 - (0.20)^3) Q = -386.73*10^-9 C total charge enclosed by the gaussian surface is q_in = q + Q = -60*10^-9 - 386.73*10^-9 q_in = 446.73*10^-9 C from gauss's law contourintegral E middot A = Vertical-bar q_in Vertical-bar /E_0 E*4*pi*b^2 = 446.73*10^-9/E_0 E = 446.73*10^-9/E_0 * 4*pi*b^2 E = 9*10^9*446.73*10^-9/(0.39)^2 E = 2.64*10^4 N/C The force on the proton due to this electric field is, F = q� * E if v be the speed of the proton in circular orbit than F = m*v^2/b so, q' * E = m*v^2/b v = squareroot (q' * E*b/m) v = squareroot [1.6*10^-19 * 2.64*10^4 *0.39/1.67*10^-27] v = 0.9931*10^6 m/s v = 9.931*10^5 m/s Let rho_max be the maximum charge density of the shell than total charge of the shell is, Q = rho_max *4/3*pi*(b^3 - a^3)

total charge on the shell is

Q =
Given q = -60 nC a = 20 cm b = 39 cm rho = -1.80 uC/m^3 total charge on the shell is Q = rho*volume = rho*4/3*phi*(b^3 - a^3) Q = -1.80*10^-6 *4/3*3.14*((0.39)^3 - (0.20)^3) Q = -386.73*10^-9 C total charge enclosed by the gaussian surface is q_in = q + Q = -60*10^-9 - 386.73*10^-9 q_in = 446.73*10^-9 C from gauss's law contourintegral E middot A = Vertical-bar q_in Vertical-bar /E_0 E*4*pi*b^2 = 446.73*10^-9/E_0 E = 446.73*10^-9/E_0 * 4*pi*b^2 E = 9*10^9*446.73*10^-9/(0.39)^2 E = 2.64*10^4 N/C The force on the proton due to this electric field is, F = q� * E if v be the speed of the proton in circular orbit than F = m*v^2/b so, q' * E = m*v^2/b v = squareroot (q' * E*b/m) v = squareroot [1.6*10^-19 * 2.64*10^4 *0.39/1.67*10^-27] v = 0.9931*10^6 m/s v = 9.931*10^5 m/s Let rho_max be the maximum charge density of the shell than total charge of the shell is, Q = rho_max *4/3*pi*(b^3 - a^3)
*volume =
Given q = -60 nC a = 20 cm b = 39 cm rho = -1.80 uC/m^3 total charge on the shell is Q = rho*volume = rho*4/3*phi*(b^3 - a^3) Q = -1.80*10^-6 *4/3*3.14*((0.39)^3 - (0.20)^3) Q = -386.73*10^-9 C total charge enclosed by the gaussian surface is q_in = q + Q = -60*10^-9 - 386.73*10^-9 q_in = 446.73*10^-9 C from gauss's law contourintegral E middot A = Vertical-bar q_in Vertical-bar /E_0 E*4*pi*b^2 = 446.73*10^-9/E_0 E = 446.73*10^-9/E_0 * 4*pi*b^2 E = 9*10^9*446.73*10^-9/(0.39)^2 E = 2.64*10^4 N/C The force on the proton due to this electric field is, F = q� * E if v be the speed of the proton in circular orbit than F = m*v^2/b so, q' * E = m*v^2/b v = squareroot (q' * E*b/m) v = squareroot [1.6*10^-19 * 2.64*10^4 *0.39/1.67*10^-27] v = 0.9931*10^6 m/s v = 9.931*10^5 m/s Let rho_max be the maximum charge density of the shell than total charge of the shell is, Q = rho_max *4/3*pi*(b^3 - a^3)
*4/3*pi*(b³ - a³)

Q = -1.80*10^-6*4/3*3.14*( (0.39)³ - (0.20)³)

Q = -386.73*10^-9 C

total charge enclosed by the gaussian surface is

q_in = q + Q = -60*10^-9 - 386.73*10^-9

q_in = 446.73*10^-9 C

from gauss's law

$\oint$ E.A = |q_in|/ $\varepsilon$ o

E*4*pi*b² = 446.73*10^-9/ $\varepsilon$ o

E = 446.73*10^-9/ $\varepsilon$ o*4*pi*b²

E = 9*10⁹*446.73*10^-9/(0.39)²

E = 2.64*10⁴ N/C

The force on the proton due to this electric field is ,

F = q'*E

if v be the speed of the proton in circular orbit than

F = m*v²/b

so,

q'*E =  m*v²/b

v = sqrt(q'*E*b/m)

v = sqrt[1.6*10^-19*2.64*10⁴*0.39/1.67*10^-27]

v = 0.9931*10⁶m/s

v = 9.931*10⁵ m/s

(b)

Let _max be themaximum charge density of the shell than total charge of the shell is,

Given q = -60 nC a = 20 cm b = 39 cm rho = -1.80 uC/m^3 total charge on the shell is Q = rho*volume = rho*4/3*phi*(b^3 - a^3) Q = -1.80*10^-6 *4/3*3.14*((0.39)^3 - (0.20)^3) Q = -386.73*10^-9 C total charge enclosed by the gaussian surface is q_in = q + Q = -60*10^-9 - 386.73*10^-9 q_in = 446.73*10^-9 C from gauss's law contourintegral E middot A = Vertical-bar q_in Vertical-bar /E_0 E*4*pi*b^2 = 446.73*10^-9/E_0 E = 446.73*10^-9/E_0 * 4*pi*b^2 E = 9*10^9*446.73*10^-9/(0.39)^2 E = 2.64*10^4 N/C The force on the proton due to this electric field is, F = q� * E if v be the speed of the proton in circular orbit than F = m*v^2/b so, q' * E = m*v^2/b v = squareroot (q' * E*b/m) v = squareroot [1.6*10^-19 * 2.64*10^4 *0.39/1.67*10^-27] v = 0.9931*10^6 m/s v = 9.931*10^5 m/s Let rho_max be the maximum charge density of the shell than total charge of the shell is, Q = rho_max *4/3*pi*(b^3 - a^3)

Q = ​_max *4/3*pi*(b³ - a³)

Given q = -60 nC a = 20 cm b = 39 cm rho = -1.80 uC/m^3 total charge on the shell is Q = rho*volume = rho*4/3*phi*(b^3 - a^3) Q = -1.80*10^-6 *4/3*3.14*((0.39)^3 - (0.20)^3) Q = -386.73*10^-9 C total charge enclosed by the gaussian surface is q_in = q + Q = -60*10^-9 - 386.73*10^-9 q_in = 446.73*10^-9 C from gauss's law contourintegral E middot A = Vertical-bar q_in Vertical-bar /E_0 E*4*pi*b^2 = 446.73*10^-9/E_0 E = 446.73*10^-9/E_0 * 4*pi*b^2 E = 9*10^9*446.73*10^-9/(0.39)^2 E = 2.64*10^4 N/C The force on the proton due to this electric field is, F = q� * E if v be the speed of the proton in circular orbit than F = m*v^2/b so, q' * E = m*v^2/b v = squareroot (q' * E*b/m) v = squareroot [1.6*10^-19 * 2.64*10^4 *0.39/1.67*10^-27] v = 0.9931*10^6 m/s v = 9.931*10^5 m/s Let rho_max be the maximum charge density of the shell than total charge of the shell is, Q = rho_max *4/3*pi*(b^3 - a^3)

Q = ​_max *4/3*3.14*((0.39)³ - (0.20)³)

Given q = -60 nC a = 20 cm b = 39 cm rho = -1.80 uC/m^3 total charge on the shell is Q = rho*volume = rho*4/3*phi*(b^3 - a^3) Q = -1.80*10^-6 *4/3*3.14*((0.39)^3 - (0.20)^3) Q = -386.73*10^-9 C total charge enclosed by the gaussian surface is q_in = q + Q = -60*10^-9 - 386.73*10^-9 q_in = 446.73*10^-9 C from gauss's law contourintegral E middot A = Vertical-bar q_in Vertical-bar /E_0 E*4*pi*b^2 = 446.73*10^-9/E_0 E = 446.73*10^-9/E_0 * 4*pi*b^2 E = 9*10^9*446.73*10^-9/(0.39)^2 E = 2.64*10^4 N/C The force on the proton due to this electric field is, F = q� * E if v be the speed of the proton in circular orbit than F = m*v^2/b so, q' * E = m*v^2/b v = squareroot (q' * E*b/m) v = squareroot [1.6*10^-19 * 2.64*10^4 *0.39/1.67*10^-27] v = 0.9931*10^6 m/s v = 9.931*10^5 m/s Let rho_max be the maximum charge density of the shell than total charge of the shell is, Q = rho_max *4/3*pi*(b^3 - a^3)

Q = ​_max *21.48*10^-2

Given q = -60 nC a = 20 cm b = 39 cm rho = -1.80 uC/m^3 total charge on the shell is Q = rho*volume = rho*4/3*phi*(b^3 - a^3) Q = -1.80*10^-6 *4/3*3.14*((0.39)^3 - (0.20)^3) Q = -386.73*10^-9 C total charge enclosed by the gaussian surface is q_in = q + Q = -60*10^-9 - 386.73*10^-9 q_in = 446.73*10^-9 C from gauss's law contourintegral E middot A = Vertical-bar q_in Vertical-bar /E_0 E*4*pi*b^2 = 446.73*10^-9/E_0 E = 446.73*10^-9/E_0 * 4*pi*b^2 E = 9*10^9*446.73*10^-9/(0.39)^2 E = 2.64*10^4 N/C The force on the proton due to this electric field is, F = q� * E if v be the speed of the proton in circular orbit than F = m*v^2/b so, q' * E = m*v^2/b v = squareroot (q' * E*b/m) v = squareroot [1.6*10^-19 * 2.64*10^4 *0.39/1.67*10^-27] v = 0.9931*10^6 m/s v = 9.931*10^5 m/s Let rho_max be the maximum charge density of the shell than total charge of the shell is, Q = rho_max *4/3*pi*(b^3 - a^3)

For maximum density charge of the shell will be just equal to the magnitude of the bead charge that is

Q = |q|

​_max *21.48*10^-2 = 60*10^-9

Given q = -60 nC a = 20 cm b = 39 cm rho = -1.80 uC/m^3 total charge on the shell is Q = rho*volume = rho*4/3*phi*(b^3 - a^3) Q = -1.80*10^-6 *4/3*3.14*((0.39)^3 - (0.20)^3) Q = -386.73*10^-9 C total charge enclosed by the gaussian surface is q_in = q + Q = -60*10^-9 - 386.73*10^-9 q_in = 446.73*10^-9 C from gauss's law contourintegral E middot A = Vertical-bar q_in Vertical-bar /E_0 E*4*pi*b^2 = 446.73*10^-9/E_0 E = 446.73*10^-9/E_0 * 4*pi*b^2 E = 9*10^9*446.73*10^-9/(0.39)^2 E = 2.64*10^4 N/C The force on the proton due to this electric field is, F = q� * E if v be the speed of the proton in circular orbit than F = m*v^2/b so, q' * E = m*v^2/b v = squareroot (q' * E*b/m) v = squareroot [1.6*10^-19 * 2.64*10^4 *0.39/1.67*10^-27] v = 0.9931*10^6 m/s v = 9.931*10^5 m/s Let rho_max be the maximum charge density of the shell than total charge of the shell is, Q = rho_max *4/3*pi*(b^3 - a^3)

​_max = 0.279*10^-6 C/m³

Given q = -60 nC a = 20 cm b = 39 cm rho = -1.80 uC/m^3 total charge on the shell is Q = rho*volume = rho*4/3*phi*(b^3 - a^3) Q = -1.80*10^-6 *4/3*3.14*((0.39)^3 - (0.20)^3) Q = -386.73*10^-9 C total charge enclosed by the gaussian surface is q_in = q + Q = -60*10^-9 - 386.73*10^-9 q_in = 446.73*10^-9 C from gauss's law contourintegral E middot A = Vertical-bar q_in Vertical-bar /E_0 E*4*pi*b^2 = 446.73*10^-9/E_0 E = 446.73*10^-9/E_0 * 4*pi*b^2 E = 9*10^9*446.73*10^-9/(0.39)^2 E = 2.64*10^4 N/C The force on the proton due to this electric field is, F = q� * E if v be the speed of the proton in circular orbit than F = m*v^2/b so, q' * E = m*v^2/b v = squareroot (q' * E*b/m) v = squareroot [1.6*10^-19 * 2.64*10^4 *0.39/1.67*10^-27] v = 0.9931*10^6 m/s v = 9.931*10^5 m/s Let rho_max be the maximum charge density of the shell than total charge of the shell is, Q = rho_max *4/3*pi*(b^3 - a^3)

Given q = -60 nC a = 20 cm b = 39 cm rho = -1.80 uC/m^3 total charge on the shell is Q = rho*volume = rho*4/3*phi*(b^3 - a^3) Q = -1.80*10^-6 *4/3*3.14*((0.39)^3 - (0.20)^3) Q = -386.73*10^-9 C total charge enclosed by the gaussian surface is q_in = q + Q = -60*10^-9 - 386.73*10^-9 q_in = 446.73*10^-9 C from gauss's law contourintegral E middot A = Vertical-bar q_in Vertical-bar /E_0 E*4*pi*b^2 = 446.73*10^-9/E_0 E = 446.73*10^-9/E_0 * 4*pi*b^2 E = 9*10^9*446.73*10^-9/(0.39)^2 E = 2.64*10^4 N/C The force on the proton due to this electric field is, F = q� * E if v be the speed of the proton in circular orbit than F = m*v^2/b so, q' * E = m*v^2/b v = squareroot (q' * E*b/m) v = squareroot [1.6*10^-19 * 2.64*10^4 *0.39/1.67*10^-27] v = 0.9931*10^6 m/s v = 9.931*10^5 m/s Let rho_max be the maximum charge density of the shell than total charge of the shell is, Q = rho_max *4/3*pi*(b^3 - a^3)
​_max = 0.279 uC/m³