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An electron, released in a region where the electric field is uniform, is observed to have an acceleration of 2.50 10 m/s in the positive x direction. Determine the magnitude of the electric field producing the acceleration.



Answer :

audreyn06

Answer: 0.14 N/C

Explanation:

The formula for the electric field is given by:

E = F/q, where:

  • E is the electric field
  • F is the net force
  • q is the charge

According to Newton's second law, the net force is also given by the equation:

F = ma, where:

  • m is the mass
  • a is the acceleration

Therefore, E = ma/q.

We are given that the particle is an electron and its acceleration is 2.5 · 10¹⁰ m/s²:

  • mass of an electron: m = 9.11 · 10⁻³¹ kg
  • charge of an electron: q = -1.6 · 10⁻¹⁹ C
  • a = 2.5 · 10¹⁰ m/s²

Let's plug these values into the formula E = ma/q:

E = (9.11 · 10⁻³¹)(2.5 · 10¹⁰ m/s²)/(-1.6 · 10⁻¹⁹)

E = -0.14 N/C

The magnitude of the electric field is 0.14 N/C.

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