Question

An infinite plane has a surface charge density of -5.0 X 10-6 C/m2. A proton is shot straight away from the plane at 2.0 X 106 m/s. How far does the proton travel before reaching its turning point?

Answer 1

Answer:

d = 0.07 m

Explanation:

Given: σ =  -5.0 × 10⁻⁶ C/m²,  v = 2.0 × 10⁶ m/s, ε = 8.85 × 10⁻¹² F/m,

Mass of Proton = 1.67 × 10⁻²⁷ kg and charge q = 1.60 × 10⁻¹⁹ C

Solution:

E = σ/2ε   (for uniform field of large single plate)

E = -5.0 X 10⁻⁶ C/m² / (2 × 8.85 X 10⁻¹² F/m)

E = 2.82 × 10 ⁵ N/C

Now Work done = K.E.

F × d = 1/2 m v²

Eq × d = 1/2 m v²

d =m v² / 2Eq

d = 1.67 × 10⁻²⁷ kg (2.0 × 10⁶ m/s)² / 2 × 2.82 × 10 ⁵ N/C × 1.60 × 10⁻¹⁹ C

d = 0.07 m

Answer 2

Answer:

the proton will travel 0.074 m before reaching its turning point.

Explanation:

Formula to calculate electric field because of the plate is as follows.

       $E = \frac{\sigma}{2 \times \epsilon_{o}}$

$= \frac{5 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = 2.82 \times 10^{5} N/C$

Now, we will consider that equilibrium of forces are present there. So,

$ma = qE$

$a = \frac{1.6 \times 10^{-19} \times 2.82 \times 10^{5}}{1.67 \times 10^{-27}} = 2.70 \times 10^{13} m/s^2$

According to the third equation of motion,

$v^{2} = 2 \times a \times d$

$d = \frac{v^{2}}{2d}$

$= \frac{(2.0 \times 10^{6})^{2}}{2 \times 2.7 \times 10^{13}} = 0.074 m$

Thus, we can conclude that the proton will travel 0.074 m before reaching its turning point.