Chapter 23: Q. 28 (page 654)

An electron traveling parallel to a uniform electric field increases its speed from $2.0 \times 10^{7} m / s to 4.0 \times 10^{7} m / s$ over a distance of $1.2 c m$ . What is the electric field strength?

Short Answer

Expert verified

The electric field strength is $E = 2.8 x 10^{5} N / C$

Step by step solution

Given information and formula used

Given :

The electron's speed ranges from : $2.0 \times 10^{7} m / s to 4.0 \times 10^{7} m / s$

Over a distance of : $1.2 c m$

Theory used :

Newton's Second law states :

$F = m \cdot a F_{E} = q E \Rightarrow q E = m \cdot a$

Formula for acceleration : ${v_{f}}^{2} = {v_{i}}^{2} + 2 a d$

Calculating the electric field strength

Using the kinematics equation, we calculate acceleration :

${v_{f}}^{2} = {v_{i}}^{2} + 2 a d \Rightarrow {v_{f}}^{2} - {v_{i}}^{2} = 2 a d \Rightarrow a = \frac{{v_{f}}^{2} - {v_{i}}^{2}}{2 d} = \frac{{(4.0 \times 10^{7})}^{2} - {(2.0 \times 10^{7})}^{2}}{2 x 0.012} = 5 \times 10^{16} m / s ²$

We can now use Newton's Second Law. The force in this situation is the electric force, which is given by $F = q E$ .

$q E = m a \Rightarrow E = \frac{m a}{q} = \frac{(9.1 x 10^{- 31}) \times (5 \times 10^{16})}{1.6 x 10^{- 19}} \Rightarrow E = 2.8 x 10^{5} N / C$

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An electron traveling parallel to a uniform electric field increases its speed from $2.0 \times 10^{7} m / s to 4.0 \times 10^{7} m / s$ over a distance of $1.2 c m$ . What is the electric field strength?

Short Answer

Step by step solution

Given information and formula used

Calculating the electric field strength

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An electron traveling parallel to a uniform electric field increases its speed from 2.0×107m/sto4.0×107m/s over a distance of 1.2cm. What is the electric field strength?

Short Answer

Step by step solution

Given information and formula used

Calculating the electric field strength

One App. One Place for Learning.

Most popular questions from this chapter

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Kinematics Physics

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Study anywhere. Anytime. Across all devices.

An electron traveling parallel to a uniform electric field increases its speed from $2.0 \times 10^{7} m / s to 4.0 \times 10^{7} m / s$ over a distance of $1.2 c m$ . What is the electric field strength?