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Chapter 22: Problem 17

How many excess electrons must be added to an isolated spherical conductor \(32.0 \mathrm{~cm}\) in diameter to produce an electric field of \(1150 \mathrm{~N} / \mathrm{C}\) just outside the surface?

Short Answer

Expert verified
About 2.04 x 10^10 excess electrons are required.

Step by step solution

01

Identify Given Values

We have a spherical conductor with a diameter of 32.0 cm, which means its radius is half of that: 16.0 cm or 0.16 m. The electric field just outside the surface is given as 1150 N/C.
02

Use Electric Field Formula for a Sphere

The electric field just outside the surface of a spherical conductor can be calculated using the formula: \[ E = \frac{k \cdot Q}{r^2} \]where \(E\) is the electric field, \(k\) is Coulomb's constant \(8.99 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\), \(Q\) is the charge, and \(r\) is the radius of the sphere.
03

Rearrange Formula to Solve for Charge \(Q\)

Rearrange the formula to find the charge \(Q\):\[ Q = \frac{E \cdot r^2}{k} \]
04

Substitute Values into the Formula

Substitute the given values into the formula:\[ Q = \frac{1150 \cdot (0.16)^2}{8.99 \times 10^9} \]Calculate this to find \(Q\).
05

Calculate \(Q\)

Perform the calculation:\[ Q = \frac{1150 \cdot 0.0256}{8.99 \times 10^9} \approx 3.27 \times 10^{-9} \text{ C} \]
06

Find the Number of Excess Electrons

The charge of one electron is approximately \(-1.6 \times 10^{-19}\) C. The number of excess electrons \(n\) can be found using \[ n = \frac{Q}{e} \], where \(e = 1.6 \times 10^{-19}\).
07

Calculate the Number of Electrons

Substitute the charge \(Q\) into the equation:\[ n = \frac{3.27 \times 10^{-9}}{1.6 \times 10^{-19}} \approx 2.04 \times 10^{10} \]This is the number of excess electrons required.

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Key Concepts

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

Coulomb's Constant
Coulomb's constant, often represented as \(k\), is a fundamental constant in physics that plays a crucial role in electrostatics. It is defined as \(8.99 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\). This constant is used to quantify the force of interaction between two point charges.
Coulomb's law states that the electric force \(F\) between two charges \(q_1\) and \(q_2\) is proportional to the product of the charges and inversely proportional to the square of the distance \(r\) between them. The formula is given by:
  • \(F = k \cdot \frac{q_1 \cdot q_2}{r^2}\)
The large value of Coulomb's constant indicates that electric forces are relatively strong compared to gravitational forces, even when the amount of charge involved is small. Understanding this constant is essential for calculations involving electric fields, as it helps determine the magnitude and direction of the force experienced by charges.
Electric Field Formula
The electric field is a vector field around charged particles. It represents the force exerted per unit charge at any point in space. For a point charge, the electric field \(E\) can be calculated using Coulomb's law. However, the formula adapts when considering geometrical shapes, such as a sphere.
For a spherical conductor, the electric field just outside the surface is given by:
  • \(E = \frac{k \cdot Q}{r^2}\)
Here, \(Q\) represents the total charge on the sphere, \(r\) is the radius, and \(k\) is Coulomb's constant. This formula derives from integrating the field contributions from all the excess electrons over the surface of the sphere.
The inverse square relationship denotes how quickly the electric field diminishes with distance, which is an essential feature of electrostatic fields. Knowing this, you can calculate the specific charge required to generate a given field strength, like in this exercise.
Excess Electrons Calculation
Calculating the number of excess electrons involves determining how many extra electrons must be added to a conductor to achieve a specific electric charge \(Q\). Each electron carries a charge of approximately \(-1.6 \times 10^{-19}\) C.
The formula to find the number of excess electrons \(n\) is:
  • \(n = \frac{Q}{e}\)
Where \(e\) represents the elementary charge, the magnitude of the charge of a single electron.
In our exercise, after determining the charge required using the electric field formula, we calculate \(n\) by dividing the total charge \(Q\) by the elementary charge. This yields the number of electrons needed to establish the prescribed electric field around the isolated spherical conductor.

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Most popular questions from this chapter

The electric field at a distance of 0.145 m from the surface of a solid insulating sphere with radius 0.355 m is 1750 N/C.(a) Assuming the sphere's charge is uniformly distributed, what is the charge density inside it? (b) Calculate the electric field inside the sphere at a distance of 0.200 m from the center.

A small conducting spherical shell with inner radius \(a\) and outer radius \(b\) is concentric with a larger conducting spherical shell with inner radius \(c\) and outer radius \(d\) (Fig. P22.47). The inner shell has total charge \(+2 q,\) and the outer shell has charge \(+4 q .\) (a) Calculate the electric field (magnitude and direction) in terms of \(q\) and the distance \(r\) from the common center of the two shells for (i) \(r < a ;\) (ii) \(a < r < b ;\) (iii) \(b < r < c ;\) (iv) \(c < r < d\) ; (v) \(r > d .\) Show your results in a graph of the radial component of \(\vec{\boldsymbol{E}}\) as a function of \(r .\) (b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (ii) inner surface of the large shell; (iv) outer surface of the large shell?

A flat sheet of paper of area 0.250 \(\mathrm{m}^{2}\) is oriented so that the normal to the sheet is at an angle of \(60^{\circ}\) to a uniform electric field of magnitude 14 \(\mathrm{N} / \mathrm{C}\) (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not? (c) For what angle \(\phi\) between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.

A point charge \(q_{1}=4.00 \mathrm{nC}\) is located on the \(x\) -axis at \(x=2.00 \mathrm{m},\) and a second point charge \(q_{2}=-6.00 \mathrm{nC}\) is on the \(y\) -axis at \(y=1.00 \mathrm{m} .\) What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius (a) \(0.500 \mathrm{m},(\mathrm{b}) 1.50 \mathrm{m},(\mathrm{c}) 2.50 \mathrm{m} ?\)

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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