Solving Charge Question: Calculating sa, sb, Er & Er

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In summary: At r = 1.5 cm, we have:E = (9x10^9)(3x10^-6)/(0.015)^2 = 3x10^7 N/CThis positive sign indicates that the electric field points away from the center of the spherical shell, which makes sense since the net charge of the shell is positive.At r = 3.5 cm, we have:E = (9x10^9)(3x10^-6)/(0.035)^2 = 2.3x10^7 N/CAt r = 8 cm, we have:E = (9x10^9)(3x10^-6)/(0.08)^2
  • #1
andrea2588
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Can anyone walk me through this? Answers will be helpful in letting me know if I am doing it right or not, but I am mostly looking for a walk through. Thanks!

I already got the ones that are zero.

A point charge of strength q1 = -2 uC is located at the center of a thick, conducting spherical shell of inner radius a = 3 cm and outer radius b = 4 cm. The conducting shell has a net charge of q2 = 3 uC.

(a) Calculate the surface charge densities on the inner (sa) and outer (sb) surfaces of the spherical shell.

sa = ? C/m2

HELP: Where on the conducting shell does the electric charge reside when equilibrium is achieved?

sb = ? C/m2

(b) Calculate the net radial electric field component at the following radii:

At r = 1.5 cm: Er = ?N/C

At r = 3.5 cm: Er = 0 N/C

At r = 8 cm: Er = ? N/C


(c) If a conducting wire is added that allows electric charge to flow between the location of q1 and the conducting shell, calculate the resulting values of the net radial components of the electric fields at the following radii:

At r = 1.5 cm: Er' = 0 N/C

At r = 3.5 cm: Er' = 0 N/C

At r = 8 cm: Er' = ? N/C
 
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  • #2


Sure, I'd be happy to walk you through this problem step by step.

First, let's start by drawing a diagram to help visualize the situation. We have a point charge (q1) located at the center of a thick, conducting spherical shell with inner radius a and outer radius b. The conducting shell has a net charge of q2.

(a) To calculate the surface charge densities on the inner and outer surfaces of the spherical shell, we need to use the formula for electric field due to a point charge, which is given by:

E = kq/r^2

where k is the Coulomb's constant (9x10^9 Nm^2/C^2), q is the charge of the point charge, and r is the distance from the point charge.

For the inner surface of the spherical shell (sa), the distance from the point charge to the inner surface is a, so we have:

E = kq1/a^2

Plugging in the values of q1 and a, we get:

E = (9x10^9)(-2x10^-6)/(0.03)^2 = -6x10^5 N/C

This negative sign indicates that the electric field points towards the center of the spherical shell, which makes sense since the point charge is negative and the inner surface of the shell is positively charged (due to the net charge of q2).

For the outer surface of the spherical shell (sb), the distance from the point charge to the outer surface is b, so we have:

E = kq1/b^2

Plugging in the values of q1 and b, we get:

E = (9x10^9)(-2x10^-6)/(0.04)^2 = -4.5x10^5 N/C

Again, the negative sign indicates that the electric field points towards the center of the spherical shell. So the surface charge densities on the inner and outer surfaces are:

sa = -6x10^5 C/m^2
sb = -4.5x10^5 C/m^2

(b) To calculate the net radial electric field component at different radii, we need to use the formula for electric field due to a charged spherical shell, which is given by:

E = kq/r^2

Since the conducting shell has a net charge of q2, we can use this formula to calculate the electric field at different
 
  • #3


As a scientist, it is important to have a clear understanding of the concepts and equations involved in solving a problem. Let's start by breaking down the problem and identifying the key information given.

We are given a point charge q1 = -2 uC located at the center of a thick, conducting spherical shell with inner radius a = 3 cm and outer radius b = 4 cm. The conducting shell has a net charge of q2 = 3 uC.

(a) To calculate the surface charge densities on the inner and outer surfaces of the spherical shell, we can use the equation:

sigma = q/A

Where sigma is the surface charge density, q is the charge, and A is the surface area.

For the inner surface, we can use the surface area of a sphere, A = 4πa^2, and the charge q2 = 3 uC. Plugging in these values, we get:

sa = (3 uC) / (4π(3 cm)^2) = 0.084 C/m^2

For the outer surface, we can use the same equation but with the outer radius b and the charge q1 = -2 uC. Plugging in these values, we get:

sb = (-2 uC) / (4π(4 cm)^2) = -0.039 C/m^2

(b) To calculate the net radial electric field component at different radii, we can use the equation:

Er = kq/r^2

Where Er is the net radial electric field, k is the Coulomb's constant, q is the charge, and r is the distance from the point charge.

At r = 1.5 cm:

Er = (9x10^9 Nm^2/C^2)(-2x10^-6 C) / (0.015 m)^2 = -8x10^6 N/C

At r = 3.5 cm:

Er = (9x10^9 Nm^2/C^2)(-2x10^-6 C) / (0.035 m)^2 = -1.63x10^6 N/C

At r = 8 cm:

Er = (9x10^9 Nm^2/C^2)(-2x10^-6 C) / (0.08 m)^2 = -281250 N/C

(c) When a conducting wire is added
 

1. What is the first step in solving a charge question?

The first step in solving a charge question is to identify the known values and the unknown value, which is typically indicated by a variable such as sa, sb, Er, or Er.

2. How do you calculate sa and sb?

Sa and sb represent the surface areas of two objects involved in a charge question. To calculate these values, you need to know the charge of each object (q1 and q2) and the distance between them (r). The formula for calculating sa is sa = q1 / (4πε0r^2), and the formula for calculating sb is sb = q2 / (4πε0r^2).

3. What is the value of ε0 and how is it used in charge questions?

ε0, also known as the permittivity of free space, is a constant value representing the electric susceptibility of a vacuum. It is used in charge questions to calculate the electric field between two charged objects, as shown in the formulas for calculating sa and sb.

4. How do you find Er in a charge question?

Er represents the relative permittivity, also known as the dielectric constant, of a material between two charged objects. To find Er, you need to know the permittivity of free space (ε0) and the permittivity of the material (εr). The formula for calculating Er is Er = εr / ε0.

5. Can you solve a charge question without knowing all of the variables?

No, you cannot solve a charge question without knowing all of the variables. Each variable, including sa, sb, Er, and Er, is necessary to calculate the electric field between two charged objects. If one variable is missing, the question cannot be fully solved.

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