How far apart are two conducting plates that have an electric field strength of 4.5 × 103V/m between them, if their potential
1 answer:
Answer:
Explanation:
Given
Electric Field Strength
Potential Difference between Plates is given by
In conducting plates a Potential difference exist between two plate which accelerate the charge when put between the conducting plates
The potential difference is given by
where E=Electric Field strength
d=distance between Plates
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Answer:
E = 3.04 10⁻⁵ N / C
Explanation:
In this problem we can use the kinematics to find how long it takes the electron to travel the plates
Let's start by reducing the magnitudes to the SI system
vₓ = 5.35 10⁶ m / s
x = 2 cm = 2 10⁻² m
y = 1 cm = 1 10⁻² m
x = vₓ t
t = x / vₓ
t = 2 10⁻² / 5.35 10⁶
t = 3,738 10⁻⁹ s
This time is also the time it takes for vertical movement to go from the center to the plate, let's look for acceleration with Newton's second law
F = m a
a = F / m = e E / m
y = + ½ a t²
= 0
We replace
y = ½ e / m E t²
E = 2 y m / e t²
Let's calculate
E = 2 1 10⁻² 9.1 10⁻³¹ / (1.6 10⁻¹⁹ 3,738 10⁻⁹)
E = 18.2 10⁻³³ / 5.98 10⁻²⁸
E = 3.04 10⁻⁵ N / C
Answer:
D) Acceleration is positive and increasing.
Explanation:
Acceleration is defined as the rate of change of velocity per unit time; in formulas:
where is the variation of velocity and
is the variation in time.
The graph shows the velocity vs the time of a moving object. We can see that is the increment on the y-axis, while
is the increment on the x axis: therefore, the ratio
is the slope of the curve. In fact, in a velocity-time graph, the slope of the curve corresponds to the acceleration of the object.
In this particular graph, we see that the slope of the curve continues to increase: therefore, the acceleration is positive (because the slope is positive, since the velocity is increasing) and increasing (because the slope is increasing).