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CALC A plastic disk of radius

R=64.0 \mathrm{~cm}

is charged on one side with ...

May 1, 2024

CALC A plastic disk of radius

R=64.0 \mathrm{~cm}

is charged on one side with a uniform surface charge density

\sigma=3.69 \mathrm{fC} / \mathrm{m}^{2}

, and then three quadrants of the disk are removed. The remaining quadrant is shown in Fig. 24.16. With

V=0

at infinity, what is the potential due to the remaining quadrant at point

P

, which is on the central axis of the original disk at distance

D=25.9 \mathrm{~cm}

from the original center? FIGURE 24.16 Problem 13.

Answer

V = \frac{\sigma R^2}{4\epsilon_0 D} \left( \sqrt{2} - 1 \right)

Solution

Calculate the electric potential due to a uniformly charged disk at a point on its axis: The electric potential

V

at a distance

D

from the center of a uniformly charged disk with surface charge density

\sigma

and radius

R

is given by the integral

V = \frac{\sigma}{2\epsilon_0} \left( \sqrt{R^2 + D^2} - D \right)

Adjust the formula for a quarter disk: Since only one quarter of the disk remains, the potential due to the quarter disk is one fourth of the potential due to the whole disk. Therefore,

V_{\text{quarter}} = \frac{1}{4} V_{\text{whole}}

Substitute the given values: Using the given values

R = 0.64 \, \text{m}

D = 0.259 \, \text{m}

, and

\sigma = 3.69 \times 10^{-15} \, \text{C/m}^2

, and the constant

\epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2

, we can calculate the potential at point

P

Perform the calculation:

V = \frac{\sigma R^2}{4\epsilon_0 D} \left( \sqrt{2} - 1 \right)

, substituting the values we get

V = \frac{3.69 \times 10^{-15} \cdot (0.64)^2}{4 \cdot 8.854 \times 10^{-12} \cdot 0.259} \left( \sqrt{2} - 1 \right)

Key Concept

Electric potential due to a uniformly charged disk

Explanation

The electric potential at a point on the axis of a charged disk is calculated by integrating the contributions from each infinitesimal charge element on the disk. For a quarter disk, the potential is a quarter of that due to the full disk.