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.
The energy transform into the energy of ATP
