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(a) How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 30.0 \(\mathrm{cm}\) in diameter to produce an electric field of 1390 \(\mathrm{N} / \mathrm{C}\) just outside the surface of the sphere? (b) What is the electric field at a point 10.0 cm outside the surface of the sphere?

Short Answer

Expert verified
(a) Approximately \(2.17 \times 10^{10}\) electrons are needed. (b) The electric field 10 cm outside is \(801.6\) N/C.

Step by step solution

01

Understand the Relationship Between Electric Field and Charge

The electric field just outside a charged sphere can be calculated using the formula \(E = \frac{kQ}{r^2}\), where \(E\) is the electric field, \(k\) is Coulomb's constant \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\), \(Q\) is the total charge, and \(r\) is the radius of the sphere.
02

Calculate the Radius of the Sphere

The sphere has a diameter of 30.0 cm, which means the radius \(r\) is half of that. Therefore, the radius is \(r = \frac{30.0}{2} = 15.0\) cm or 0.15 meters.
03

Calculate the Total Charge Required

Rearranging the formula \(E = \frac{kQ}{r^2}\) to find the charge \(Q\), we get \(Q = \frac{Er^2}{k}\). Given \(E = 1390\) N/C, \(r = 0.15\) m, substitute these values and \(k\) into the equation: \(Q = \frac{1390 \times (0.15)^2}{8.99 \times 10^9}\). Calculating this gives us \(Q \approx 3.48 \times 10^{-9}\) C.
04

Determine the Number of Excess Electrons

One electron has a charge of approximately \(1.602 \times 10^{-19}\) C. To find the number of excess electrons, divide the total charge by the charge of one electron: \(\frac{3.48 \times 10^{-9}}{1.602 \times 10^{-19}} \approx 2.17 \times 10^{10}\) electrons.
05

Calculate the Electric Field 10 cm Outside the Sphere

To find the electric field at a point 10.0 cm outside the surface, use the same formula for the electric field with the new distance. The total distance from the center is \(r' = 0.15 + 0.10 = 0.25\) m. Substitute \(r'\) back into the equation: \(E' = \frac{kQ}{(0.25)^2}\). Using the charge from Step 3: \(E' = \frac{8.99 \times 10^9 \times 3.48 \times 10^{-9}}{0.0625}\). Calculate this to get \(E' \approx 801.6\) N/C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's law
Coulomb's law is fundamental in understanding how electric forces act between charges. This law tells us that the electric force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. The equation representing this law is:
  • \( F = \frac{k |q_1 q_2|}{r^2} \)
Here, \( F \) is the force between the charges, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, \( r \) is the distance between the centers of the two charges, and \( k \) is Coulomb's constant, \( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \).
This principle is not only used to calculate forces but also to understand how an electric field is generated around charges. By applying this law, one can derive the formula to find the electric field created by a charge at a certain distance, making it central to problems involving electric fields around spherical charges.
Charge distribution
In physics, understanding charge distribution is key to analyzing electric fields. Charge distribution refers to how electric charge is arranged over a volume, an area, or a line. When charges are distributed uniformly, it means the charge per unit area or volume remains constant throughout.
For a uniformly charged sphere, the total charge \( Q \) affects the electric field just outside its surface. The formula used to compute such a scenario takes into account the radius of the sphere:
  • \( E = \frac{kQ}{r^2} \)
Here, \( E \) represents the electric field strength at a point just outside the surface of the sphere, and \( r \) is the distance from the center of the sphere to the point where the electric field is measured.
This uniform distribution simplifies calculations because the electric field just outside a spherical shell of charge behaves as if all charge is concentrated at the center.
Excess electrons
Excess electrons in a system indicate a surplus of negatively charged particles. In electrostatic problems, these excess electrons contribute to the formation of net charge. One determines the total charge, \( Q \), by multiplying the number of excess electrons by the charge of a single electron:
  • Charge of an electron: \( 1.602 \times 10^{-19} \) C
To find the number of electrons creating a specific charge \( Q \), divide \( Q \) by the charge of one electron.
For instance, if the goal is to have a particular electric field around a sphere by distributing electrons evenly, one first calculates the total charge required and then divides by the elementary charge of an electron to get the number of excess electrons.
This concept helps bridge the microscopic world of electrons with the macroscopic effects of electric fields in electrostatic problems.
Calculation of electric field
Calculating the electric field involves understanding how a charge influences the space around it. The electric field \( E \) at any point gives the force per unit charge experienced by a small positive test charge placed at that point.
  • Formula: \( E = \frac{kQ}{r^2} \)
In the exercise, the electric field just outside a sphere is determined using this formula. The known value of \( k \), charge \( Q \), and radius \( r \) allows us to compute \( E \).
When moving away from the surface, the point where you calculate the field changes, requiring an updated distance \( r' \) from the center of the sphere. This change involves incorporating the additional distance in calculations to find the new electric field.
Through such calculations, students explore how varying distances from a charge can either increase or decrease the electric field's strength, illustrating the inverse square relationship inherent in electric field problems.

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Most popular questions from this chapter

A solid metal sphere with radius 0.450 \(\mathrm{m}\) carries a net charge of 0.250 \(\mathrm{nC}\) . Find the magnitude of the electric field (a) at a point 0.100 \(\mathrm{m}\) outside the surface of the sphere and (b) at a point inside the sphere, 0.100 \(\mathrm{m}\) below the surface.

The gravitational force between two point masses separated by a distance \(r\) is proportional to \(1 / r^{2},\) just like the electric force between two point charges. Because of this similarity between gravitational and electric interactions, there is also a Gauss's law for gravitation. (a) Let \(\vec{g}\) be the acceleration due to gravity caused by a point mass \(m\) at the origin, so that \(\vec{g}=-\left(G m / r^{2}\right) \hat{r}\) . Consider a spherical Gaussian surface with radius \(r\) centered on this point mass, and show that the flux of \(\vec{g}\) through this surface is given by $$\oint \vec{g} \cdot d \vec{A}=-4 \pi G m$$ (b) By following the same logical steps used in Section 22.3 to obtain Gauss's law for the electric field, show that the flux of \(\vec{g}\) through any closed surface is given by $$\oint \vec{g} \cdot d \vec{A}=-4 \pi G M_{\mathrm{encl}}$$ where \(M_{\text { encl }}\) is the total mass enclosed within the closed surface.

A point charge of \(-2.00 \mu C\) is located in the center of a spherical cavity of radius 6.50 \(\mathrm{cm}\) inside an insulating charged solid. The charge density in the solid is \(\rho=7.35 \times 10^{-4} \mathrm{C} / \mathrm{m}^{3}\) Calculate the electric field inside the solid at a distance of 9.50 \(\mathrm{cm}\) from the center of the cavity.

A hollow, conducting sphere with an outer radius of 0.250 \(\mathrm{m}\) and an inner radius of 0.200 \(\mathrm{m}\) has a uniform surface charge density of \(+6.37 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) A charge of \(-0.500 \mu \mathrm{C}\) is now introduced into the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere. (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere?

An insulating sphere of radius \(R=0.160 \mathrm{m}\) has uniform charge density \(\rho=+7.20 \times 10^{-9} \mathrm{C} / \mathrm{m}^{3} .\) A small object that can be treated as a point charge is released from rest just outside the surface of the sphere. The small object has positive charge \(q=3.40 \times 10^{-6} \mathrm{C}\) . How much work does the electric field of the sphere do on the object as the object moves to a point very far from the sphere?

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