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lord [1]
2 years ago
11

A small charge q is placed near a large spherical charge Q. The force experienced by both charges is F. The electric eld created

by Q at the position of q is:
Physics
1 answer:
anastassius [24]2 years ago
7 0

The electric field created by Q at the position of q is \frac{F}{Q}.

The given parameters:

  • <em>Magnitude of charge, = q</em>
  • <em>Spherical charge, = Q</em>
  • <em>Force experienced by both charges, =  F</em>

The electric field created by Q at the position of q is calculated as follows;

E = \frac{F}{Q} \\\\

where;

  • <em>E is the magnitude of electric field strength </em>
  • <em>F is the force experienced by both charges</em>
  • <em>Q is the charge</em>

Thus, the electric field created by Q at the position of q is \frac{F}{Q}.

Learn more about electric field here: brainly.com/question/14372078

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Answer:

   t_total = 23.757 s

Explanation:

This is a kinematics exercise.

Let's start by calculating the distance and has to reach the limit speed of

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in this time he runs

         y₁ = v₀ t₁ + ½ a t₁²

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         y₁ = 73.627 m

This is the time and distance traveled until reaching the maximum speed, which will be constant throughout the rest of the trip.

           x_total = x₁ + x₂

           x₂ = x_total - x₁

           x₂ = 373 - 73,627

           x₂ = 299.373 m

this distance travels at constant speed,

           v = x₂ / t₂

           t₂ = x₂ / v

           t₂ = 299.373 / 18.8

           t₂ = 15.92 s

therefore the total travel time is

           t_total = t₁ + t₂

           t_total = 7.833 + 15.92

           t_total = 23.757 s

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