Question

We find 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 = 2.9 m from a disk of radius R = 2.9 cm that has a total charge of 5.0 µC using the exact formula as follows. (Enter your answer to at least one decimal place.)
E = 2πK σ(1- [y/(R² + y²)]) j^

Answer 1

Answer:

7. 01 * 10^7 N/C

Explanation:

Parameters given:

Distance, y = 2.9 m

Radius, R = 2.9 cm = 0.029m

Charge, Q = 5.0 µC = 5 * 10^(-6) C

Given that:

E = 2πKσ(1- [y/(R² + y²)]) j^

Charge density, σ, is given as:

σ = Q/A = Q/πR²

=> E = 2πKQ/πR² (1- [y/(R² + y²)]) j^

E = 2KQ/R² (1 - [y/(R² + y²)]) j^

E = (2 * 9 * 10^9 * 5 * 10^(-6) / 0.029²) (1 - [2.9/(0.029² + 2.9²)]) j^

E = 107015.46 * 10^3 *(1 - 2.9/8.41) j^

E = 107015.46 * 10^3 * 0.655 j^

E = 7.01 * 10^7 j^ N/C

The magnitude of the electric field is 7.01 * 10^7 N/C. The j^ shows the direction.