Chapter 4: Q30E (page 779)

An infinitely long line of charge has linear charge density $5.00 x 10^{- 12} C / m$ . A proton (mass $1.67 {x 10}^{- 27} k g$ , charge $1.67 {x 10}^{- 19} k g$ ) is 18.0 cm from the line and moving directly toward the line at $3.50 {x 0}^{3} m l s$ . (a) Calculate the proton’s initial kinetic energy. (b) How close does the proton get to the line of charge?

Short Answer

Expert verified

a) The initial kinetic energy of protons is $1 \times 10 - 20 J$

b) It is 9cm close.

Step by step solution

Kinetic Energy

It is defined as thework needed to accelerate a body of a given mass from rest to its statedvelocity.

$K = \frac{1}{2} m v^{2}$

The proton’s initial kinetic energy

Given;

$\begin{array}{l} A = 5.0 \times 10^{- 12} C / m \\ m = 1.67 \times 10 - 27 kg \\ v_{i} = 3.5 \times 10^{3} m l s \\ r_{i} = 18 cm \end{array}$

a) Therefore, the kinetic energy is;

$\begin{array}{l} K_{i} - \frac{m_{p}^{} {V_{i}}^{2}}{2} \\ = \frac{1.67 \times 10^{- 27} (3.5 \times 10^{3})}{2} = 1 \times 10 - 00] \end{array}$

Hence, the initial kinetic energy of protons is

How close does the proton get to the line of charge

b) The kinetic energy and the potential energy are;

$\begin{array}{l} K_{t} + u_{1} = K_{f} + u_{f} \\ V_{f} - V_{i} = \frac{K_{i}}{e} \end{array}$

By substituting;

$v_{f} - V_{i} = \frac{1 \times 10^{- 20}}{1.6 \times 10^{- 19}} = 0.0625 V$

Therefore;

$V_{i} - V_{f} = \frac{λ}{2 π €_{o}} I n \frac{r_{f}}{r_{i}}$

By substituting;

$I n \frac{r_{f}}{r_{i}} = \frac{- 0.0625 \times 2 \times π \times 8.85 \times 10^{- 12}}{5 \times 10^{- 12}} = - 0.69$

So $r_{f}$ is;

$\begin{array}{l} r_{f} = r_{l}^{} e x p (- 0.69} \\ 18 e x g (- 0.69) \\ = 9 cm \end{array}$

Hence, it is 9cm close.

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Short Answer

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Kinetic Energy

The proton’s initial kinetic energy

How close does the proton get to the line of charge

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