Question

In Example 24.6, we found that the electric field of a charged disk approaches that of a charged particle for distances y that are large compared to R, the radius of the disk. To see a numerical instance of this, calculate the magnitude of the electric field a distance y = 3.1 m from a disk of radius R = 3.1 cm that has a total charge of 6.8 µC using the exact formula as follows. (Enter your answer to at least one decimal place.) E = 2πkσ ( 1 − y/sqrt(R2 + y2)) j

Answer 1

The electric field is calculated by the formula,

$E = 2 \pi k \sigma \left( 1- \frac{y}{\sqrt{y^2 + R^2}} \right ) = \frac{2 \pi \sigma}{4 \pi \epsilon_0} \left( 1- \frac{y}{\sqrt{y^2 + R^2}} \right )$

Now we have to calculate the charge density which is,

$\sigma = \frac{q}{\pi R^2} = \frac{6.8 \times 10^{-6}}{\pi (3.1 \times 10^{-2})^2} = 2.25 \times 10^{-3} \, C/m^2$

Substituting the values we get,

$\Rightarrow E = \frac{2 \pi (2.25 \times 10^{-3})}{4 \pi \epsilon_0} \left( 1- \frac{3.1}{\sqrt{(3.1)^2 + (3.1 \times 10^{-2})^2}} \right )$

$\Rightarrow E = \frac{(2.25 \times 10^{-3})}{2 \epsilon_0} \left( 1- \frac{3.1}{\sqrt{(3.1)^2 + (3.1 \times 10^{-2})^2}} \right )$

$\Rightarrow E = 6352.45 \, N/C$