An air-track glider attached to a spring oscillates between the 14.0 cm mark and the 71.0 cm mark on the track. The glider compl
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Answer:
Explanation:
Given that:
Charge (q) on the particle = 3 × 10⁻⁸ C
mass (m) of the particle = 6 × 10⁻⁹ kg
at a distance x = 15 cm , the velocity in the plate = 900 m/s²
For the square plate, the surface charged density σ = -8 × 10⁻⁶ C/m²
To start with calculating the electric field as a result of the square plate; we use the formula;
On the square plate; The electric force F = Eq
The acceleration
For the particle, the velocity at distance x = 7 m can be calculated by using the formula:
From the calculation, we realize that the charge acting between the particle and the plate is said to be "opposite".
Hence, the force is an attractive force.
Similarly, there is a gradual increase exhibited by the velocity of the particle.
Therefore, the particles get to the detector, but the detector failed to get detect due to the velocity which is greater than 1000 m/s.