Question:

A small charged particle with a mass of 2.5 x 10^-6 kg and a charge of +1.6 x 10^-19 C is released in an electric field. The particle accelerates horizontally with an acceleration of 5.0 x 10^6 m/s^2. The electric field intensity and direction are unknown.

a) Determine the magnitude of the electric field intensity. b) Determine the direction of the electric field.

Assume negligible air resistance and use appropriate constants for all calculations.

Answer:

To solve this problem, we will use the equation for the force experienced by a charged particle in an electric field:

F = qE

Where:

• F is the force experienced by the charged particle (in newtons)
• q is the charge of the particle (in coulombs)
• E is the electric field intensity (in newtons per coulomb)

Also, we know that force is equal to mass times acceleration:

F = ma

Given data:

• mass (m) = 2.5 x 10^-6 kg
• charge (q) = +1.6 x 10^-19 C
• acceleration (a) = 5.0 x 10^6 m/s^2

a) Determining the magnitude of the electric field intensity:

Using the equation F = ma and substituting qE for F, we can set up the following equation:

qE = ma

Rearranging the equation to isolate E:

E = ma / q

Plugging in the given values:

E = (2.5 x 10^-6 kg) * (5.0 x 10^6 m/s^2) / (1.6 x 10^-19 C)

Calculating:

E ≈ 7.8 x 10^11 N/C

Therefore, the magnitude of the electric field intensity is approximately 7.8 x 10^11 N/C.

b) Determining the direction of the electric field:

The direction of the electric field can be determined from the direction of the acceleration experienced by the charged particle. According to the given information, the particle accelerates horizontally.

Since the force experienced by a positive charge is in the same direction as the electric field, the direction of the electric field is also horizontal.

Hence, the direction of the electric field is horizontal.