   Chapter 16, Problem 10P

Chapter
Section
Textbook Problem

On planet Tehar, the free-fall acceleration is the same as that on the Earth, but there is also a strong downward electric field that is uniform close to the planet’s surface. A 2.00-kg ball having a charge of 5.00 μC is thrown upward at a speed of 20.1 m/s. It hits the ground after an interval of 4.10 s. What is the potential difference between the starting point and the top point of the trajectory?

To determine
The potential difference.

Explanation

Given info: Mass of the ball (m) is 2.00 kg. The charge on the ball (q) is 5.00μC . Speed (v) at which the ball is thrown is 20.1 m/s. The time interval (t) is 4.10 s, Mass of the ball (m) is 2.00 kg. The charge on the ball (q) is 5.00μC . Speed (v) at which the ball is thrown is 20.1 m/s. The time interval (t) is 4.10 s.

Explanation:

Formula to calculate the displacement is,

Δy=v22a

• a is the acceleration.

The acceleration is given by,

a=2vt

From the above equations,

Δy=vt4

Substitute 20.1 m/s for v and 4.10 s for t.

Δy=(20.1m/s)(4.10s)4=20.6m

Formula to calculate the electric field is,

E=m(a+g)q

• g is the acceleration due to gravity.

The acceleration is given by,

a=2vt

From the above equations,

E=m(g2vt)q

Substitute 2

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