7. Protons are accelerated by an electric field between metal plates A and B, in a vacuum - Scottish Highers Physics - Question 7 - 2022

To find the kinetic energy (KE) of the proton, we use the formula:

KE = rac{1}{2}mv^2

Where:

• m = mass of proton = $1.67 imes 10^{-27}$ kg
• v = speed of proton at plate A = $3.8 imes 10^5$ m/s

Calculating the kinetic energy:

KE = rac{1}{2} (1.67 imes 10^{-27} ext{ kg}) (3.8 imes 10^{5} ext{ m/s})^2 $KE = 1.2 imes 10^{-16} ext{ J}$

The work done (W) on the proton can be calculated from the potential difference (V) using the formula:

$W = qV$

Where:

• q = charge of proton = $1.6 imes 10^{-19}$ C
• V = potential difference between the plates = $2.8 imes 10^3$ V

Calculating the work done:

$W = (1.6 imes 10^{-19} ext{ C}) (2.8 imes 10^3 ext{ V})$ $W = 4.48 imes 10^{-16} ext{ J}$

At plate B, the kinetic energy will be the sum of the initial kinetic energy and the work done:

$KE_B = KE_A + W$

Calculating the kinetic energy at plate B:

$KE_B = 1.2 imes 10^{-16} ext{ J} + 4.48 imes 10^{-16} ext{ J} = 5.68 imes 10^{-16} ext{ J}$

Setting this equal to the kinetic energy formula and solving for speed:

KE_B = rac{1}{2}mv_B^2 5.68 imes 10^{-16} = rac{1}{2}(1.67 imes 10^{-27})v_B^2

Solving for $v_B$:

There will be no effect on the speed of the proton at plate B. Although the distance between plates A and B is doubled, the potential difference remains unchanged. Therefore, the work done on the proton remains the same, resulting in the same gain in kinetic energy. Thus, the final speed of the proton at plate B will remain unaffected.