A capacitor is constructed from two conducting with a plate separation of d. The voltage between the plates is 31.2 volts and the electric field between the two plates is measured to be E = 2805 V/m.
What is the minimum amount of force needed to push an electron from the positive plate back to the negative plate?
F=eE=1.6•10⁻¹⁹•2805 = 4.5•10⁻¹⁶ N
To calculate the minimum amount of force needed to push an electron from the positive plate back to the negative plate of a capacitor, we need to consider the electric force acting on the electron.
The electric force (F) on an electron in an electric field (E) can be calculated using the formula:
F = q * E
where q is the charge of the electron. The charge of an electron is given as -1.6 x 10^-19 coulombs.
In this case, the electric field (E) between the two plates of the capacitor is given as 2805 V/m.
So, let's plug in the values:
F = (-1.6 x 10^-19 C) * (2805 V/m)
Now we can calculate the force needed:
F = -4.48 x 10^-17 N
Therefore, the minimum amount of force needed to push an electron from the positive plate back to the negative plate of the capacitor is -4.48 x 10^-17 Newtons. It should be noted that the negative sign indicates that the force is in the opposite direction of the electric field.