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Chapter 16: Problem 12

How many electrons must be removed from each of two 5.0 -kg copper spheres to make the electric force of repulsion between them equal in magnitude to the gravitational attraction between them?

Short Answer

Expert verified
Answer: Approximately 9.64 脳 10^10 electrons must be removed from each sphere.

Step by step solution

01

Write down the relevant formulas.

First, we write down the formulas for the gravitational force (F_gravity) and the electric force (F_electric). F_gravity = G * (m1 * m2) / r^2 F_electric = k * (q1 * q2) / r^2 where G is the gravitational constant (6.674 脳 10^-11 N m^2/kg^2), m1 and m2 are the masses of the copper spheres, k is the electrostatic constant (8.99 脳 10^9 N m^2/C^2), q1 and q2 are the charges on the copper spheres, and r is the distance between the centers of the spheres. We are given that the spheres have a mass of 5.0 kg each, so m1 = m2 = 5.0 kg, and we want the forces to have equal magnitudes: F_gravity = F_electric.
02

Equate the magnitudes of the forces and solve for the charge.

To find the number of electrons, we first need to find the charge needed on each sphere. We equate the magnitudes of the gravitational and electric forces and solve for q1 * q2: G * (m1 * m2) / r^2 = k * (q1 * q2) / r^2 By cancelling out r^2 and rearranging the equation, we get: q1 * q2 = G * (m1 * m2) / k Since the charges on the copper spheres have the same magnitude (q1 = q2), we can substitute q^2 for q1 * q2: q^2 = G * (m1 * m2) / k Now we can solve for q: q = sqrt(G * (m1 * m2) / k)
03

Calculate the value of the charge.

By plugging the given values and constants into the equation, we can calculate the charge: q = sqrt((6.674 脳 10^-11 N m^2/kg^2) * (5.0 kg * 5.0 kg) / (8.99 脳 10^9 N m^2/C^2)) q 鈮� 1.54 脳 10^-8 C
04

Calculate the number of electrons.

Now we know the total charge needed on each sphere, we can find the number of electrons that must be removed. The charge of an individual electron is 1.60 脳 10^-19 C. We divide the charge on each sphere by the charge of an electron to find the number of electrons: number of electrons = q / e number of electrons = (1.54 脳 10^-8 C) / (1.60 脳 10^-19 C) number of electrons 鈮� 9.63 脳 10^10 Since we cannot have a fraction of an electron, we will round up to the nearest whole number: Number of electrons to be removed from each sphere 鈮� 9.64 脳 10^10

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Most popular questions from this chapter

A coaxial cable consists of a wire of radius \(a\) surrounded by a thin metal cylindrical shell of radius \(b\). The wire has a uniform linear charge density \(\lambda>0\) and the outer shell has a uniform linear charge density \(-\lambda\). (a) Sketch the field lines for this cable. (b) Find expressions for the magnitude of the electric field in the regions \(r \leq a, a
(a) Use Gauss's law to prove that the electric field outside any spherically symmetric charge distribution is the same as if all of the charge were concentrated into a point charge. (b) Now use Gauss's law to prove that the electric field inside a spherically symmetric charge distribution is zero if none of the charge is at a distance from the center less than that of the point where we determine the field.
A parallel-plate capacitor consists of two flat metal plates of area \(A\) separated by a small distance \(d\). The plates are given equal and opposite net charges \(\pm q\) (a) Sketch the field lines and use your sketch to explain why almost all of the charge is on the inner surfaces of the plates. (b) Use Gauss's law to show that the electric field between the plates and away from the edges is $E=q /\left(\epsilon_{0} A\right)=\sigma / \epsilon_{0} \cdot(\mathrm{c})$ Does this agree with or contra- dict the result of Problem \(70 ?\) Explain. (d) Use the principle of superposition and the result of Problem 69 to arrive at this same answer. [Hint: The inner surfaces of the two plates are thin, flat sheets of charge.]
Sketch the electric field lines near two isolated and equal (a) positive point charges and (b) negative point charges. Include arrowheads to show the field directions.
A thin, flat sheet of charge has a uniform surface charge density \(\sigma(\sigma / 2\) on each side). (a) Sketch the field lines due to the sheet. (b) Sketch the field lines for an infinitely large sheet with the same charge density. (c) For the infinite sheet, how does the field strength depend on the distance from the sheet? [Hint: Refer to your field line sketch.J (d) For points close to the finite sheet and far from its edges, can the sheet be approximated by an infinitely large sheet? [Hint: Again, refer to the field line sketches.] (e) Use Gauss's law to show that the magnitude of the electric field near a sheet of uniform charge density \(\sigma\) is $E=\sigma /\left(2 \epsilon_{0}\right)$
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