Question

Electric potential at a point is V = Ar\(^3\) + B. Find charge enclosed in a sphere of radius 1m, centered at r = 0

• A. \(-4\pi\epsilon_{0}A\)
• B. \(-8\pi\epsilon_{0}A\)
• C. \(-12\pi\epsilon_{0}A\)
• D. \(-16\pi\epsilon_{0}A\)

Hint:

The relationship between E and V (\(E = -dV/dr\)) and Gauss's Law are fundamental concepts. For spherically symmetric charge distributions, applying Gauss's law with a spherical Gaussian surface is the most efficient method. Be careful with the vector dot product \(\vec{E} \cdot d\vec{S}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the electric potential as a function of radial distance, V(r). We need to find the total charge enclosed within a sphere of radius 1m using this potential function.
Step 2: Key Formula or Approach:
1. Find the electric field \(\vec{E}\) from the electric potential \(V\) using the relation \(\vec{E} = -\frac{dV}{dr}\hat{r}\) for a spherically symmetric potential.
2. Use Gauss's Law to find the enclosed charge \(q_{in}\). Gauss's Law states that the net electric flux \(\Phi_E\) through a closed surface is equal to the enclosed charge divided by \(\epsilon_0\).
\[ \Phi_E = \oint \vec{E} \cdot d\vec{S} = \frac{q_{in}}{\epsilon_0} \] Step 3: Detailed Explanation:
Step 1: Find the Electric Field (E)
The given potential is \(V = Ar^3 + B\).
The electric field is the negative gradient of the potential. In this case, it is:
\[ E = -\frac{dV}{dr} = -\frac{d}{dr}(Ar^3 + B) \] \[ E = -3Ar^2 \] The negative sign indicates that for a positive A, the field is directed radially inward.
Step 2: Apply Gauss's Law
Consider a Gaussian surface as a sphere of radius r. The electric field is uniform in magnitude on this surface and is directed radially. The area vector \(d\vec{S}\) is also directed radially outward.
The electric flux \(\Phi_E\) through the sphere of radius r is:
\[ \Phi_E = \oint \vec{E} \cdot d\vec{S} = E \cdot (Area) = E \cdot (4\pi r^2) \] Note: Since \(\vec{E}\) is inward and \(d\vec{S}\) is outward, \(\vec{E} \cdot d\vec{S} = EdS \cos(180^\circ) = -EdS\). \[ \Phi_E = \oint (-E) dS = -E \oint dS = -E (4\pi r^2) \] Substituting the expression for E: \[ \Phi_E = -(-3Ar^2)(4\pi r^2) = 12\pi Ar^4 \] Let's re-evaluate. Using vector notation: \(\vec{E} = -3Ar^2 \hat{r}\).
\[ \Phi_E = \oint (-3Ar^2 \hat{r}) \cdot (dS \hat{r}) = \int -3Ar^2 dS = -3Ar^2 \int dS = -3Ar^2 (4\pi r^2) = -12\pi Ar^4 \] This seems to have a unit mismatch. Let's restart the flux calculation. The field at radius r is \(E(r) = -3Ar^2\). The flux through a sphere of radius r is \(E(r) \times (\text{Surface Area})\). \[ \Phi_E(r) = E(r) \times 4\pi r^2 = (-3Ar^2) \times (4\pi r^2) \] This cannot be correct. Let's use the differential form: \(\rho = \epsilon_0 \nabla \cdot \vec{E}\). Or stick to the integral form. \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} \] \[ \vec{E} = - \nabla V = - \frac{\partial V}{\partial r} \hat{r} = -3Ar^2 \hat{r} \] \[ \Phi_E = \int_{\text{sphere}} (-3Ar^2 \hat{r}) \cdot (r^2 \sin\theta d\theta d\phi \hat{r}) \] The surface element is \(d\vec{A} = r^2 \sin\theta d\theta d\phi \hat{r}\). We need to evaluate at \(r=1\). \[ \Phi_E = \int_0^{2\pi} \int_0^{\pi} (-3A(1)^2 \hat{r}) \cdot (\hat{r} (1)^2 \sin\theta d\theta d\phi) \] \[ \Phi_E = -3A \int_0^{2\pi} d\phi \int_0^{\pi} \sin\theta d\theta = -3A (2\pi) [-\cos\theta]_0^{\pi} = -6\pi A (1 - (-1)) = -12\pi A \] This flux calculation is correct. From Gauss's law: \(q_{in} = \epsilon_0 \Phi_E\).
\[ q_{in} = \epsilon_0 (-12\pi A) = -12\pi\epsilon_0 A \] This is for a sphere of radius r=1m. Step 4: Final Answer:
The charge enclosed in the sphere is \(-12\pi\epsilon_0 A\).