6. A charged particle remains stationary in an upwardly directed E field between two parallel charged plates separated by 2 cm. Compute the potential difference V between
the plates if the particle has a mass of 4 x 10-13 kg and a charge of +2.4 x 10-18 C.
M g = Q E = Q V/d
where d is the plate separation and Q is the charge.
Solve for d. You know all the other terms
To compute the potential difference (V) between the plates, we need to use the concept of electric potential energy (U) and the formula U = qV, where q is the charge of the particle and V is the potential difference.
In this case, the charged particle remains stationary, which means the electric potential energy is equal to zero. This occurs when the electrical force (Fe) exerted on the particle is balanced by the gravitational force (Fg) acting in the opposite direction.
The electrical force on the particle is given by Fe = qE, where E is the magnitude of the electric field between the plates.
The gravitational force acting on the particle is given by Fg = mg, where m is the mass of the particle and g is the acceleration due to gravity.
Since the particle remains stationary, we have Fe = Fg.
Substituting the respective formulas, we have qE = mg.
Rearranging the equation to solve for E, we get E = (mg)/q.
Now we can compute the electric field (E) between the plates.
E = (mg)/q
= (4 x 10^(-13) kg) x (9.8 m/s^2) / (2.4 x 10^(-18) C)
≈ 6.47 x 10^4 N/C
Next, we can calculate the potential difference (V) between the plates.
V = E x d
= (6.47 x 10^4 N/C) x (0.02 m)
= 1.294 x 10^3 V
Therefore, the potential difference between the plates is approximately 1.294 x 10^3 volts.