Given below are two statements:
Statement I: An electric dipole is placed at the centre of a hollow

Q: Why is the electric field not zero on the sphere in Statement I if the flux is zero?

Flux is the *net* number of field lines passing through the surface. For a dipole, the number of field lines exiting the sphere equals the number entering it, making the net flux zero. However, the electric field itself is the vector sum of the fields from the two charges, which does not cancel out to zero at the surface.

Q: Doesn't Gauss's Law say $\oint \vec{E} \cdot d\vec{A} = 0$ implies $E = 0$?

No! $\oint \vec{E} \cdot d\vec{A} = 0$ only implies that the *integral* is zero. $E$ can be non-zero as long as the positive contributions to the integral perfectly balance the negative contributions.

Q: Why does all the charge reside on the surface of a solid metallic sphere?

In a conductor, charges are free to move. Because like charges repel, they push each other as far apart as possible, which means they migrate to the outermost surface until electrostatic equilibrium is reached.

Q: What if the sphere in Statement II was a solid non-conducting sphere?

If it were a solid non-conducting sphere with uniform charge density, the charge would be distributed throughout the volume. In that case, a Gaussian surface inside would enclose some charge, making both the electric field and the flux non-zero.

Q: Is the electric field exactly zero everywhere inside the hollow sphere in Statement I?

No, the electric field is non-zero everywhere inside the hollow sphere (except possibly at some specific points depending on the exact geometry, but generally non-zero) because the dipole creates a field throughout the space.

Question

Given below are two statements:
Statement I An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere.
Statement II If $R$ is the radius of a solid metallic sphere and $Q$ be the total charge on it. The electric field at any point on the spherical surface of radius $r$ ($r < R$) is zero but the electric flux passing through this closed spherical surface of radius $r$ is not zero.
In the light of the above statements, choose the correct answer from the options given below.

Options:

(A) Both Statement I and Statement II are true.

(B) Statement I is true but Statement II is false.

(C) Both Statement I and Statement II are false.

(D) Statement I is false but Statement II is true.

Sigma Percentile · Accepted Answer

### Demystifying Gauss's Law: Dipoles and Conductors

Gauss's Law is one of the most elegant and powerful tools in electrostatics. It relates the electric flux flowing out of a closed surface to the charge enclosed within it. However, it is also a breeding ground for conceptual traps. Let's dissect this problem statement by statement to uncover the physical realities behind the math.

### Analyzing Statement I: The Dipole in a Hollow Sphere

Statement I asks us to imagine an electric dipole placed at the center of a hollow sphere. A dipole consists of two charges of equal magnitude but opposite sign, $+q$ and $-q$, separated by a small distance.

According to Gauss's Law, the total electric flux $\Phi$ through any closed surface is given by:

$$\Phi_{net} = \frac{q_{enc}}{\varepsilon_0}$$

Since the sphere encloses the entire dipole, the net enclosed charge is simply $q_{enc} = +q - q = 0$. Therefore, the net electric flux through the sphere is absolutely zero.

But does zero flux imply that the electric field $\vec{E}$ is zero everywhere on the surface? **Absolutely not!** Flux is a measure of the *net* flow of field lines. For a dipole, the number of field lines exiting the sphere equals the number entering it. However, the electric field at any specific point on the sphere is the vector sum of the fields from both charges. Because the charges are spatially separated, their individual fields do not perfectly cancel out everywhere on the surface. Thus, $\vec{E} 
eq 0$ everywhere.

This makes Statement I perfectly **True**.

### Analyzing Statement II: The Solid Metallic Sphere

Statement II introduces a solid **metallic** sphere of radius $R$ carrying a total charge $Q$. The word "metallic" is the critical key here. In any conductor in electrostatic equilibrium, charges are free to move. Because like charges repel, they push each other as far apart as possible, migrating entirely to the outer surface.

This means that inside the bulk of the metal, there is absolutely no excess charge. If we construct a spherical Gaussian surface of radius $r < R$ completely inside the metal, the charge enclosed by this imaginary surface is exactly zero ($q_{enc} = 0$).

Applying Gauss's Law to this inner surface:

$$\Phi = \oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0} = 0$$

So, the electric flux through this inner surface is zero. Furthermore, the electric field inside a conductor in electrostatic equilibrium is always zero ($\vec{E} = 0$).

Statement II correctly states that the electric field is zero, but it incorrectly claims that the electric flux is non-zero. Because the flux is actually zero, Statement II is **False**.

*(Note: A common pitfall is confusing a solid metallic sphere with a solid non-conducting sphere. If the sphere were non-conducting with a uniform volume charge density, there would be charge enclosed within $r < R$, making both the field and the flux non-zero. Always read the material properties carefully!)*

### Final Conclusion

By carefully applying Gauss's Law and the properties of conductors, we have established that Statement I is True and Statement II is False. This leads us directly to the correct option, (b).

Select Answer:

Visualized Solution

Analyzing Statement I

Gauss's Law for the Sphere

Electric Field of a Dipole

Analyzing Statement II

Gaussian Surface Inside the Conductor

Flux and Field Inside the Conductor

Final Conclusion

The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

Demystifying Gauss's Law

Analyzing Statement I

Analyzing Statement II

Final Conclusion

Similar Questions

A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

A point charge $+ Q$ is placed just outside an imaginary hemispherical surface of radius $R$ as shown in the figure. Which of the following statements is/are correct?

Let $ρ (r) = \frac{Q}{π R ^{4}} r$ be the charge density distribution for a solid sphere of radius $R$ and total charge $Q$ . For a point $P$ inside the sphere at distance $r_{1}$ from the centre of the sphere, the magnitude of electric field is

Which of the following statement(s) is/are correct?

A non-conducting solid sphere of radius $R$ is uniformly charged. The magnitude of the electric field due to the sphere at a distance $r$ from its centre

If the electric flux entering and leaving an enclosed surface respectively is $ϕ_{1}$ and $ϕ_{2}$ , the electric charge inside the surface will be

Select Answer:

Visualized Solution

Analyzing Statement I

Gauss's Law for the Sphere

Electric Field of a Dipole

Analyzing Statement II

Gaussian Surface Inside the Conductor

Flux and Field Inside the Conductor

Final Conclusion

The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

Demystifying Gauss's Law

Analyzing Statement I

Analyzing Statement II

Final Conclusion

Similar Questions

A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct?

Which of the following statement(s) is/are correct?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its centre

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the electric flux entering and leaving an enclosed surface respectively is ϕ1​ and ϕ2​, the electric charge inside the surface will be

A point charge $+ Q$ is placed just outside an imaginary hemispherical surface of radius $R$ as shown in the figure. Which of the following statements is/are correct?

A non-conducting solid sphere of radius $R$ is uniformly charged. The magnitude of the electric field due to the sphere at a distance $r$ from its centre

If the electric flux entering and leaving an enclosed surface respectively is $ϕ_{1}$ and $ϕ_{2}$ , the electric charge inside the surface will be