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A proton (mass = 1.67 × 1027 kg, charge = 1.60 × 10-1 C) moves from point A to



point B under the influence of an electrostatic force only. At point A the proton moves with a speed of 60 km/s. At point B the speed of the proton is 80 km/s.



Determine the potential difference Vb - Va •



a.



-33 volt

b.



-20 volt

c.



-15 volt

d.



+15 volt

e.


  • 33 volt

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Verified Answer :

e3van22

To determine the potential difference \( V_B - V_A \), we can use the relationship between work done by the electrostatic force and the change in electric potential energy.

The work done by the electrostatic force on the proton as it moves from point A to point B is equal to the change in its kinetic energy:

\[ W = \Delta KE \]

We can express the change in kinetic energy as:

\[ \Delta KE = \frac{1}{2} m v_B^2 - \frac{1}{2} m v_A^2 \]

Substituting the given values:

\[ \Delta KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times (80 \times 10^3)^2 - \frac{1}{2} \times 1.67 \times 10^{-27} \times (60 \times 10^3)^2 \]

\[ \Delta KE \approx \frac{1}{2} \times 1.67 \times 10^{-27} \times (6400 \times 10^6 - 3600 \times 10^6) \]

\[ \Delta KE \approx \frac{1}{2} \times 1.67 \times 10^{-27} \times 2800 \times 10^6 \]

\[ \Delta KE \approx 2.34 \times 10^{-21} \text{ joules} \]

The work done by the electrostatic force is equal to the change in electric potential energy, which can be expressed as the change in potential multiplied by the charge of the proton:

\[ W = q \times (V_B - V_A) \]

Solving for \( V_B - V_A \):

\[ V_B - V_A = \frac{W}{q} \]

\[ V_B - V_A = \frac{2.34 \times 10^{-21}}{1.60 \times 10^{-19}} \]

\[ V_B - V_A \approx -14.625 \text{ volts} \]

Rounded to the nearest volt, the potential difference \( V_B - V_A \) is approximately \( -15 \) volts.

Therefore, the correct answer is option c. \( -15 \) volt.