A parallel plate capacitor is made of two 3.0 cm times 3.0 cm square plates that are separated by...

Question:

A parallel plate capacitor is made of two 3.0 cm times 3.0 cm square plates that are separated by 1.0 mm are connected to a 12.0 V battery. If and electron is released from the negative electrode, what speed will it reach by the time it arrives at the positive electrode?

Conservation of Mechanical Energy

A charged particle kept inside an electric field gains electrostatic potential energy, and this electrostatic potential energy get converted into kinetic energy when the charged particle is free to move. This statement is known as the conservation of mechanical energy.

Data Given

• Area of a plate of a parallel plate capacitor {eq}A = 3 \ \rm cm \times 3 \ \rm cm {/eq}
• Separation between the plates {eq}d = 1 \ \rm mm = 10^{-3} \ \rm m {/eq}
• Potential applied across the plates {eq}V = 12 \ \rm V {/eq}

Using the conservation of mechanical energy, the electrostatic potential energy of the electron at the negative plate converts to kinetic energy as it moves towards the positive plate, so

{eq}\begin{align} KE = EPE \end{align} {/eq}

{eq}\begin{align} \frac{1}{2}mv^2 = qV \end{align} {/eq}

{eq}\begin{align} v =\sqrt{\frac{2 qV}{m}} \end{align} {/eq}

{eq}\begin{align} v =\sqrt{\frac{2 \times 1.6 \times 10^{-19} \ \rm C \times 12.0 \ \rm V}{9.11 \times 10^{-31} \ \rm kg}} \end{align} {/eq}

{eq}\begin{align} \color{blue}{\boxed{ \ v = \ 2.05 \times 10^6 \ \rm m/s \ }} \end{align} {/eq}