Math Problem Statement
A charged conducting spherical shell of radius R = 5 m with total charge q = 31 μC produces the electric field given by
E⃗ (r)={014πϵ0qr2r^forforr<Rr>R , )
Enter an expression for the electric potential inside the sphere ( r < R ) in terms of the given quantities, assuming the potential is zero at infinity.
-First take the dot product, then integrate what you have left. -Integrate from infinity (where the potential will be zero) to r. -Since the electric field is discontinuous, you will have to break the integration into two parts, one from infinity to R and then the other from R to r.
Solution
The electric potential at a point inside the conducting spherical shell, with , can be determined by integrating the electric field. The electric field is given as:
0 & \text{for } r < R \\ \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \hat{r} & \text{for } r \ge R \end{cases}$$ Since the electric field is zero for $$r < R$$, we focus on finding the electric potential inside the spherical shell, starting from infinity (where the potential is zero). ### Step 1: Express the relationship between the electric potential and electric field. The electric potential $$V(r)$$ is related to the electric field $$\vec{E}(r)$$ through: $$dV = -\vec{E} \cdot d\vec{r}$$ Since the electric field is radially symmetric, the dot product simplifies as: $$dV = -E(r) \, dr$$ ### Step 2: Break the integral into two parts We now integrate the electric field in two parts: 1. From infinity to $$R$$ (where the potential is zero at infinity). 2. From $$R$$ to $$r$$ (where $$r < R$$). Since the electric field is zero for $$r < R$$, the potential will remain constant inside the shell, and the contribution to the potential inside the shell will come from the integral from infinity to $$R$$. ### Part 1: Integral from infinity to $$R$$ For $$r \ge R$$, the electric field is $$\vec{E}(r) = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}$$, so: $$V(R) - V(\infty) = - \int_{\infty}^{R} \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \, dr$$ Since $$V(\infty) = 0$$, this simplifies to: $$V(R) = \frac{1}{4 \pi \epsilon_0} \frac{q}{R}$$ ### Part 2: Integral from $$R$$ to $$r$$ (for $$r < R$$) Inside the conducting shell, $$\vec{E}(r) = 0$$, meaning there is no electric field to contribute to the potential. Therefore, the potential is constant inside the shell and equal to the potential at $$r = R$$, so: $$V(r) = V(R) = \frac{1}{4 \pi \epsilon_0} \frac{q}{R} \quad \text{for} \quad r < R$$ ### Final Answer: The electric potential $$V(r)$$ inside the spherical shell is: $$V(r) = \frac{1}{4 \pi \epsilon_0} \frac{q}{R} \quad \text{for} \quad r < R$$ Where: - $$q = 31 \, \mu C = 31 \times 10^{-6} \, C$$ - $$R = 5 \, m$$ Would you like to see a detailed breakdown of the integral steps or any clarification on the concept? Here are some related questions you might find useful: 1. Why is the electric field inside a conducting spherical shell zero? 2. How does the electric potential behave outside a spherical shell with a charge? 3. What happens to the electric field if the shell were not conducting? 4. How do we find the electric potential at points outside the shell? 5. What role does the concept of equipotential surfaces play in this problem? **Tip**: In cases involving spherical symmetry, remember that the electric field at a point inside a conducting shell is always zero, which results in a constant potential inside the shell.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Electric Potential
Electric Field
Integration
Conducting Spherical Shell
Formulas
Electric Potential Relation: dV = -E(r) * dr
V(R) = (1 / 4πϵ₀) * q / R
Theorems
Gauss's Law
Electric Field Inside a Conducting Shell is Zero
Suitable Grade Level
Grades 11-12
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