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Chapter 22: Problem 17

A dipole is completely enclosed by a spherical surface. Describe how the total electric flux through this surface varies with the strength of the dipole.

Short Answer

Expert verified
Answer: The total electric flux through a spherical surface enclosing a dipole does not vary with the strength of the dipole. The electric flux will always be zero, irrespective of the dipole strength, because the positive and negative charges collectively have no net charge enclosed within the surface.

Step by step solution

01

Understand the dipole

A dipole consists of two charges, one positive and one negative, separated by a small distance. The strength of a dipole is defined by the dipole moment, p, which is the product of the charge and the separation distance. Mathematically, we can write the dipole moment as p = qd, where q is the charge and d is the separation distance between the charges.
02

Apply Gauss's law

Gauss's law states that the total electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space (蔚鈧�). Mathematically, Gauss's law can be written as 桅 = Q_enclosed / 蔚鈧�.
03

Calculate the enclosed charge

Now, since we have a dipole enclosed by the spherical surface, the total enclosed charge Q_enclosed is the sum of the positive charge and the negative charge within the surface. As a dipole consists of an equal amount of positive and negative charges, the total enclosed charge within the sphere is zero. Therefore, Q_enclosed = 0.
04

Determine the total electric flux

Using Gauss's law, we can now determine the total electric flux through the spherical surface. Since the enclosed charge is zero, 桅 = Q_enclosed / 蔚鈧� = 0 / 蔚鈧� = 0. This means that the total electric flux through the surface is zero, regardless of the strength of the dipole enclosed by the surface.
05

Interpret the result

Thus, the total electric flux through the spherical surface enclosing a dipole does not vary with the strength of the dipole. The electric flux will always be zero, irrespective of the dipole strength, because the positive and negative charges collectively have no net charge enclosed within the surface.

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Most popular questions from this chapter

A long conducting wire with charge distribution \(\lambda\) and radius \(r\) produces an electric field of \(2.73 \mathrm{~N} / \mathrm{C}\) just outside the surface of the wire. What is the magnitude of the electric field just outside the surface of another wire with charge distribution \(0.81 \lambda\) and radius \(6.5 r ?\)

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