Question

A spherical ball having a uniformly distributed charge Q and radius R pulsates with frequency \(\omega\) such that the radius changes by \(\pm 10%\), as shown in the figure below. Which of the following is(are) correct?

• A. The net outward electric flux across a spherical surface of radius \(r>1.5R\) pulsates with a frequency \(\omega\)
• B. The net outward electric flux across a spherical surface of radius \(r = 2R\) is \(\frac{Q}{\epsilon_0}\)
• C. The potential fluctuates with frequency \(\omega\) at \(r=2R\)
• D. The electric field inside the sphere at \(r=0.5R\) will not be time dependent

Hint:

Gauss's Law is a very powerful tool, especially for symmetric charge distributions. Remember its key insight: the flux through a closed surface depends *only* on the total charge enclosed, not on how that charge is distributed within the surface or on the motion of the charge inside, as long as the surface itself isn't moving in a way that changes the enclosed charge.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves the application of Gauss's Law to a spherically symmetric charge distribution that is changing in size over time. We need to analyze how the electric flux, electric field, and electric potential behave both inside and outside this pulsating sphere.
Step 2: Key Formula or Approach:
1. Gauss's Law: The net electric flux \(\Phi_E\) through a closed surface is equal to the total charge enclosed (\(Q_{\text{enc}}\)) divided by the permittivity of free space (\(\epsilon_0\)). \[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \] 2. Electric Field of a Spherical Charge Distribution: For a point outside a spherically symmetric charge distribution (at distance \(r\) from the center), the electric field is the same as that of a point charge Q located at the center: \(E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}\). Inside the distribution, the field depends on the charge enclosed within radius \(r\). 3. Electric Potential: The potential at a distance \(r\) from the center (outside the sphere) is \(V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}\).
Step 3: Detailed Explanation:
(A) The net outward electric flux across a spherical surface of radius \(r>1.5R\) pulsates with a frequency \(\omega\).
Let the Gaussian surface be a sphere of radius \(r\). The radius of the charged ball is \(R(t)\), which pulsates. The maximum radius is \(1.1R\) and the minimum is \(0.9R\). The condition \(r>1.5R\) ensures that the Gaussian surface is always outside the pulsating charged ball. According to Gauss's Law, the flux depends only on the enclosed charge, \(\Phi_E = Q_{\text{enc}}/\epsilon_0\). Since the Gaussian surface always encloses the entire charge Q, the enclosed charge is constant and equal to Q. Therefore, the net outward electric flux is constant and does not pulsate. Statement (A) is incorrect.
(B) The net outward electric flux across a spherical surface of radius \(r = 2R\) is \(\frac{Q}{\epsilon_0}\).
Similar to the reasoning for (A), a spherical surface of radius \(r=2R\) is always outside the pulsating sphere (since its maximum radius is \(1.1R\)). Therefore, this surface always encloses the total charge Q. By Gauss's Law, the net outward electric flux is \(\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{Q}{\epsilon_0}\). This value is constant. Statement (B) is correct.
(C) The potential fluctuates with frequency \(\omega\) at \(r=2R\).
For any point outside a spherically symmetric charge distribution, the electric potential is given by \(V(r) = \frac{Q}{4\pi\epsilon_0 r}\). The point of observation is at a fixed radius \(r=2R\), and the total charge Q is constant. The potential at this fixed point depends only on Q and r, neither of which is changing. The pulsation of the sphere's radius \(R(t)\) does not affect the potential at a fixed point outside the sphere. Therefore, the potential does not fluctuate. Statement (C) is incorrect.
\textit{Note: This is true in electrostatics. If we consider radiation due to accelerating charges (as the pulsating surface charges are accelerating), there would be an electromagnetic wave, and the potential would become more complex (Liénard-Wiechert potential). However, for a typical electrostatics problem context, we assume the quasi-static approximation holds, where the fields adjust instantaneously. In this approximation, C is incorrect.}
(D) The electric field inside the sphere at \(r=0.5R\) will not be time dependent.
Let the time-varying radius of the sphere be \(R'(t)\). The charge Q is uniformly distributed throughout the volume of the sphere. The charge density is \(\rho(t) = \frac{Q}{\frac{4}{3}\pi (R'(t))^3}\).
Consider a point at a fixed radius \(r_0 = 0.5R\). The radius of the sphere \(R'(t)\) varies between \(0.9R\) and \(1.1R\). Thus, the point \(r_0=0.5R\) is always inside the sphere.
Using Gauss's Law for a point inside the sphere, the electric field is due to the charge enclosed within radius \(r_0\): \[ E(r_0, t) = \frac{Q_{\text{enc}}(r_0, t)}{4\pi\epsilon_0 r_0^2} \] The enclosed charge is \(Q_{\text{enc}}(r_0, t) = \rho(t) \times \frac{4}{3}\pi r_0^3 = \frac{Q}{\frac{4}{3}\pi (R'(t))^3} \times \frac{4}{3}\pi r_0^3 = Q \frac{r_0^3}{(R'(t))^3}\).
So, the electric field is: \[ E(r_0, t) = \frac{1}{4\pi\epsilon_0 r_0^2} \left( Q \frac{r_0^3}{(R'(t))^3} \right) = \frac{Q r_0}{4\pi\epsilon_0 (R'(t))^3} \] Since \(R'(t)\) is pulsating with frequency \(\omega\), the electric field \(E\) at the fixed point \(r_0 = 0.5R\) is time-dependent. It fluctuates as \((R'(t))^{-3}\). Statement (D) is incorrect.
Step 4: Final Answer:
Based on the analysis, only statement (B) is correct.