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A particle with a charge of -5.0 micro C and a mass of 4.3 x 10^{-6} kg is released from rest at...

Question:

A particle with a charge of {eq}\rm -5.0 \ \mu C {/eq} and a mass of {eq}\rm 4.3 \times 10^{-6} \ kg {/eq} is released from rest at point A and accelerates toward point B, arriving there with a speed of 18 m/s. The only force acting on the particle is the electric force. What is the potential difference between A and B?

Conservation of Mechanical Energy

As we know that electrostatic force is a conservative force, so for a charged particle in an electric field, the total mechanical energy of the system remains conserved, i.e. change in the potential energy of the system is equal to the change in the kinetic energy of the system.

{eq}\begin{align} \Delta KE = \Delta PE \end{align} {/eq}

Answer and Explanation:

Data Given

  • Charge on the particle {eq}q = 5 \ \rm \mu C = 5 \times 10^{-6} \ \rm C {/eq}
  • Mass of the particle {eq}m = 4.3 \times 10^{-6} \ \rm kg {/eq}
  • Speed of the particle at A {eq}v_i = 0 \ \rm m/s {/eq}
  • Speed of the particle at B {eq}v_f = 18 \ \rm m/s {/eq}

Using the conservation of mechanical energy

{eq}\begin{align} \Delta KE = \Delta PE \end{align} {/eq}

{eq}\begin{align} KE_B - KE_A = q. \Delta V \end{align} {/eq}

{eq}\begin{align} \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = q. \Delta V \end{align} {/eq}

{eq}\begin{align} \Delta V = \frac{m (v_f^2 - v_i^2)}{2q} \end{align} {/eq}

{eq}\begin{align} \Delta V = \frac{ 4.3 \times 10^{-6} \ \rm kg \times ((18 \ \rm m/s )^2 - (0 \ \rm m/s)^2)}{2 \times 5 \times 10^{-6} \ \rm C } \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{ \ \Delta V \ = \ 139.3 \ \rm V}} \end{align} {/eq}


Learn more about this topic:

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Energy Conservation and Energy Efficiency: Examples and Differences

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