Question

A hollow cylindrical shell of length L and radius R has charge Q uniformly distributed along its length. What is the electric potential at the center of the cylinder?

Accepted Answer

Step 1: Recognize that the electric potential at a point is the sum of contributions from all charge elements. For this problem, the charge is uniformly distributed along the length of the cylindrical shell, so we need to integrate over the length of the cylinder to find the total potential. Step 2: Write the expression for the electric potential due to a small charge element dq at a distance r from the center. The formula for the electric potential is $$ V = \frac{1}{4 \pi \epsilon_0} \frac{dq}{r} $$, where $$ \epsilon_0  is $$the permittivity of free space. Step 3: Express dq in terms of the linear charge density $$ \lambda $$, where $$ \lambda = \frac{Q}{L} . $$Since the charge is uniformly distributed, $$ dq = \lambda dx $$, where dx is a small segment of the cylinder's length. Step 4: Set up the integral to calculate the total potential at the center of the cylinder. The distance from any charge element on the shell to the center is equal to the radius R of the cylinder. Substitute $$ dq = \lambda dx $$ and $$ r = R $$ into the potential formula: $$ V = \int_{0}^{L} \frac{1}{4 \pi \epsilon_0} \frac{\lambda dx}{R} . $$Step 5: Simplify the integral. Since $$ \lambda $$, $$ R $$, and $$ \frac{1}{4 \pi \epsilon_0} $$ are constants, they can be factored out of the integral. The integral becomes $$ V = \frac{\lambda}{4 \pi \epsilon_0 R} \int_{0}^{L} dx . $$Evaluate the integral $$ \int_{0}^{L} dx $$, which equals L, and substitute $$ \lambda = \frac{Q}{L}  to $$express the final formula for the electric potential.

A hollow cylindrical shell of length L and radius R has charge Q uniformly distributed along its length. What is the electric potential at the center of the cylinder?

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Key Concepts

Electric Potential

Cylindrical Symmetry

Gauss's Law