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Norma-Jean [14]
4 years ago
9

I need the answer right now pls help me

Physics
1 answer:
yaroslaw [1]4 years ago
7 0

The AREA of the shaded region is the moving object's displacement.

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We find that the electric field of a charged disk approaches that of a charged particle for distances y that are large compared
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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.

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Did the assignment.

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