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3 months ago
10

You are working on a laboratory device that includes a small sphere with a large electric charge Q. Because of this charged sphe

re, there is a strong electric field surrounding your device. Other researchers in your laboratory are complaining that your electric field is affecting their equipment. You think about how you can obtain the large electric field that you need close to the sphere but prohibit the field from reaching your colleagues. You decide to surround your device with a spherical transparent plastic shell. The nonconducting shell is given a uniform charge distribution. Required:a. The shell is placed so that the small sphere is at the exact center of the shell. Determine the charge that must be placed on the shell to completely eliminate the electric field outside of the shell. b. What if the shell moves? Does the small sphere have to be at the center of the shell for this scheme to work?
Physics
1 answer:
Yuliya22 [3.3K]3 months ago
5 0

Response:

The primary consequence is an increase in induced charge at the nearest points. However, the overall net charge remains zero, meaning it does not influence the flow.

We can utilize Gauss's law to solve this problem

      Ф = ∫ e. dA = q_{int} / ε₀

The flow of the field is directly correlated to the charge within it. Consequently, placing a Gaussian surface beyond the non-conductive spherical shell means the flow will be zero since the sphere’s charge equals the charge induced in the shell, resulting in a net charge of zero. This evaluation shows that the shell effectively obstructs the electric field.

According to Gauss's law, if the sphere is offset, the only effect it generates is an increment in induced charge at the nearest points. Nevertheless, the net charge remains zero, so it does not impact the flow; irrespective of the sphere's position, the total induced charge is consistently equal to the charge on the sphere.

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Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells
ValentinkaMS [3465]

Answer:

a) Blood mass is m= 5.7876kg

b) The count of blood cells is  N_t=1.04*10^{13}

Explanation:

From the problem statement, we learn that

         The blood volume is  V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3

         The density of blood is  \rho_b = 1060 kg/m^3

         % of blood which consists of cells is  = 45.0%

        the % of blood that is  plasma is  = 55.0%

        density of blood cells is  \rho_d = 1125kg/m^3

         % of cells that are white is  = 1%

        % of cells that are red is  = 99%

         The red blood cell diameter is  = 7.5 \mu m = 7.5*10^{-6}m

         The red blood cell radius is  = \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m

The mass is generally represented mathematically as

               m = \rho_b * V_b

Substituting values

            m = 1060 * 0.00546

               m= 5.7876kg

Cell mass is m_c = 45% of m

                         = 0.45 * 5,7876

                         = 2.60442 kg

The volume of cells is V_c = \frac{m_c}{\rho_d}

                                      = \frac{2.60442}{1125}

                                      = 2.315 *10^{-3} m^3

The white blood cells volume is V_w = 1% of the cells volume

                                                         = \frac{1}{100} * 2.315*10^{-3}

                                                       = 2.315*10^{-5}m^3

The volume of a single cell is V_s = 4 \pi r^3

                                                                        = 4*(3.142) * (3.75*10^{-6})^3

                                                                        = 2.21*10^{-16}m^3

The red blood cells volume is V_r = V_c - V_w

                                                           =2.315*10^{-3} - 2.315*10^{-5}

                                                           = 2.29*10^{-3}m^3

The total red blood cell count is  = \frac{V_r}{V_s}

                                                     = \frac{2.29 *10^{-3}}{2.21*10^{-16}}

                                                    = 1.037*10^{13}

The total white blood cell count is   =\frac{V_w}{V_s}

                                                          = \frac{2.315 * 10^{-5}}{2.21*10^{-16}}

                                                          = 1.04*10^{11}

The overall number of blood cells is  N_t= 1.037*10^{13} + 1.04*10^{11}

                                                        N_t=1.04*10^{13}

6 0
2 months ago
Read 2 more answers
Jo, Daniel and Helen are pulling a metal ring. Jo pulls with a force of 100N in one direction and Daniel with a force of 140N in
inna [3103]

Answer:

She exerts a force of 40 N.

Explanation:

The fact that the ring remains stationary indicates that the forces are in equilibrium.

Let’s denote Jo's force as x.

The equation to consider is

140 = x + 100

x = 40

5 0
3 months ago
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