Business, 15.02.2020 03:19 rbalexander25

A point charge q1 = -9.6 μC is located at the center of a thick conducting spherical shell of inner radius a = 2.4 cm and outer radius b = 4.6 cm, The conducting spherical shell has a net charge of q2 = 1.5 μC.

1) What is Ex(P), the value of the x-component of the electric field at point P, located a distance 8.4 cm along the x-axis from q1?

2) What is Ey(P), the value of the y-component of the electric field at point P, located a distance 8.4 cm along the x-axis from q1?

3) What is Ex(R), the value of the x-component of the electric field at point R, located a distance 1.2 cm along the y-axis from q1?

4) What is Ey(R), the value of the y-component of the electric field at point R, located a distance 1.2 cm along the y-axis from q1?

5) What is σb, the surface charge density at the outer edge of the shell?

6) For how many values of x: (4.6 cm < x < infinity) is it true that Ex = 0?

a)none

b)one

c)more than one

7) Define E2 to be equal to the magnitude of the electic field at r = 1.2 cm when the charge on the outer shell (q2) is equal to 1.5 μC. Define Eo to be equal to the magnitude of the electric field at r = 1.2 cm if the charge on the outer shell (q2) were changed to 0. Compare E2 and Eo.

a)E2 < Eo

b)E2 = Eo

c)E2 > Eo

Answers: 2

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Answers

Answer from: reecedstceklein

Part 1: The value of x component of electric field at point P is $-1.03 \times 10^7 \frac{ N}{ C}$

Part 2: The value of y component of electric field at point P is 0.

Part 3: The value of x component of electric field at point R is 0.

Part 4: The value of y component of electric field at point R is $-5.06 \times 10^8 \frac{ N}{ C}$ .

Part 5: The value of surface density at the outer edge of the shell is $-3.83 \times 10^{-4} C/m^2$ .

Part 6: None ,The field is treated as if it is a single point charge outside the conducting wall and there after extends to infinity diminishing by a rate of r^2 .

Part 7:The fields are equal as the charge on the outer shell does not affect the field on within the shell. ( E_2=E_o )

Explanation:

Part 1

$E = k \frac{ Q}{ r^2}$ is the Electric field of a point charge

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

a = 2.4 cm = 0.024 m is inner radius of the conducting shell

b = 4.1 cm = 0.041 m is outer radius of the conducting shell

x= 8.4 cm = 0.084 m is the location of P on x

So the value is given as

$E_x( P) = k \left(\frac{ q_1 + q_2}{ x^2}\right) \\E_x( P) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0.084^2}\right) \\= -1.03 \times 10^7 \frac{ N}{ C}$

The value of x component of electric field at point P is $-1.03 \times 10^7 \frac{ N}{ C}$

Part 2

$E = k \frac{ Q}{ r^2}$ is the Electric field of a point charge

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

a = 2.4 cm = 0.024 m is inner radius of the conducting shell

b = 4.1 cm = 0.041 m is outer radius of the conducting shell

y = 0 cm = 0.0 m is the location of P on y

So the value is given as

$E_y( P) = k \left(\frac{ q_1 + q_2}{ y^2}\right) \\E_y( P) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0}\right) \\= 0$

The value of y component of electric field at point P is 0.

Part 3

$E = k \frac{ Q}{ r^2}$ is the Electric field of a point charge

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

a = 2.4 cm = 0.024 m is inner radius of the conducting shell

b = 4.1 cm = 0.041 m is outer radius of the conducting shell

x = 0 cm = 0.0 m is the location of R on x

So the value is given as

$E_x( R) = k \left(\frac{ q_1 + q_2}{ x^2}\right) \\E_x( R) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0}\right) \\= 0$

The value of x component of electric field at point R is 0.

Part 4

$E = k \frac{ Q}{ r^2}$ is the Electric field of a point charge

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

a = 2.4 cm = 0.024 m is inner radius of the conducting shell

b = 4.1 cm = 0.041 m is outer radius of the conducting shell

y =1.2 cm = 0.012 m is the location of R on y

So the value is given as

$E_y( R) = k \left(\frac{ q_1 + q_2}{ y^2}\right) \\E_y( R) = 9 \times 10^9 \left(\frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{0.012}\right) \\=-5.06 \times 10^{8} N/C$

The value of y component of electric field at point R is $-5.06 \times 10^8 \frac{ N}{ C}$ .

Part 5

$\sigma = \frac{ q_{enclosed}}{ 4 \pi r^2}$ is the Surface charge density

From given data

$q_1 = -9.6 \mu C$ is point charge at the center

$q_2 = 1.5\mu C$ is net charge of the conducting shell

a = 2.4 cm = 0.024 m is inner radius of the conducting shell

b = 4.1 cm = 0.041 m is outer radius of the conducting shell

So the value is given as

$\sigma_b = \frac{ q_{enclosed}}{ 4 \pi b^2}\\\sigma_b = \frac{ -9.6 \times 10^{-6}+ 1.5 \times 10^{-6}}{ 4 \pi 0.041^2}\\\sigma_b = -3.83 \times 10^{-4} C/m^2$

The value of surface density at the outer edge of the shell is $-3.83 \times 10^{-4} C/m^2$ .

Part 6

None because the field is treated as if it is a single point charge outside the conducting wall and there after extends to infinity diminishing by a rate of r^2 .

Part 7

The fields are equal as the charge on the outer shell does not affect the field on within the shell. ( E_2=E_o )

Answer from: Quest

The answer may be c ..

Answer from: Quest

answer; test-retest;

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