The potential difference between points A and B is -2.44 ✕ 10^-3 V. The negative sign indicates that the potential decreases from point A to point B.
Part 1: X-component of the electric field (Ex)
Net force acting on the molecule: The electric force is the only force acting on the molecule as it slows down. We can express this with Newton's second law: F_net = ma = eEx
where:
m is the mass of the nitrogen molecule (4.65 ✕ 10−26 kg)
a is the acceleration of the molecule (negative since it's slowing down)
e is the charge of the molecule (+e)
Ex is the x-component of the electric field
Acceleration: The initial velocity is 1.72 ✕ 103 m/s, and the final velocity is 0 m/s.
The distance traveled is 0.841 mm. We can use the following equation to find the acceleration:
v^2 = u^2 + 2ad
Rearranging for a:
a = (v^2 - u^2) / 2*d
a = (-1.72 ✕ 10^3 m/s)^2 / (2 * 0.841 ✕ 10^-3 m)
a = -2.45 ✕ 10^7 m/s^2
Solving for Ex:
| F_net | = eEx
| ma | = e*Ex
Ex = |m*a| / e
Ex = | 4.65 ✕ 10^-26 kg * -2.45 ✕ 10^7 m/s^2 | / 1.60 ✕ 10^-19 C
Ex = 6.07 ✕ 10^3 N/C
Therefore, the x-component of the electric field is 6.07 ✕ 10^3 N/C in the negative x-direction.
Part 2: Potential Difference (VAB)
Work done by the electric field: The work done by the electric field on the molecule is equal to the change in its kinetic energy. As the molecule comes to a stop, its kinetic energy decreases to zero.
W = ∆KE = -1/2 * m * v^2
W = -1/2 * 4.65 ✕ 10^-26 kg * (1.72 ✕ 10^3 m/s)^2
W = -3.91 ✕ 10^-16 J
Potential difference: Work done by an electric field is related to the potential difference by the equation:
W = q * VAB
where:
q is the charge of the particle (+e)
VAB is the potential difference between points A and B
Solving for VAB:
VAB = W / q
VAB = -3.91 ✕ 10^-16 J / (+1.60 ✕ 10^-19 C)
VAB = -2.44 ✕ 10^-3 V
Therefore, the potential difference between points A and B is -2.44 ✕ 10^-3 V. The negative sign indicates that the potential decreases from point A to point B.