note: E is for exponetial

two large horizontal, parallel metal plates are 2.5cm apart in air and the upper is maintained at a positive potential relative to the lower so that the field strenght between them is 2.1x10E3vmE-1

the potential difference between plates is 5.25x10E-1volts

Ques: If an electron of charge 1.8x10E-19C and mass 9.8x10E-31kg is liberated from rest at lower plate. find its speed on reaching the upper plate.

i got 4.3x10ms as answer

pls never sure if am right with this answer getting tough time with it.

The energy the electron receives is Vq, or 5.25E-1*1.6E-19 joules, and that is equal to 1/2 m v^2

v=sqrt 2(energy)/9,8E-31

I don't get your answer.

8.57*10^-31

Well, it seems like you're having a tough time with this question. But don't worry, I'm here to help... or at least make you smile!

Let's break it down:

To find the speed of the electron on reaching the upper plate, we need to use the concept of potential energy and kinetic energy.

The potential difference between the plates is given as 5.25x10E-1 volts. Since the electron is moving from the lower plate to the upper plate, it will experience a decrease in potential energy.

The equation for potential energy is:

Potential Energy = Charge x Potential Difference

Plugging in the values, we get:

Potential Energy = (1.8x10E-19 C) x (5.25x10E-1 V)

Now, since we know the potential energy, we can use the conservation of energy to find the kinetic energy:

Potential Energy = Kinetic Energy

Kinetic Energy = (1/2) x Mass x Velocity^2

Rearranging the equation, we get:

Velocity^2 = (2 x Potential Energy) / Mass

Plugging in the values and solving for velocity, we find:

Velocity = square root of [(2 x (1.8x10E-19 C) x (5.25x10E-1 V)) / (9.8x10E-31 kg)]

Now, you can calculate the final answer. But remember, even if you make a mistake, it's all a learning experience. So don't be too hard on yourself! Keep smiling and keep going!

To find the speed of the electron on reaching the upper plate, you can use the principle of conservation of energy.

Step 1: Find the electric potential energy gained by the electron as it moves from the lower plate to the upper plate.

The electric potential difference (V) between the plates is given as 5.25x10^-1 volts. The charge of the electron (q) is given as 1.8x10^-19 C. Therefore, the electric potential energy gained by the electron is:

Potential Energy (PE) = q * V
= (1.8x10^-19 C) * (5.25x10^-1 V)

Step 2: Find the kinetic energy of the electron when it reaches the upper plate.

The kinetic energy (KE) of an object is given by the formula:

KE = (1/2) * m * v^2

where m is the mass of the object and v is its velocity.

Given: mass (m) of the electron = 9.8x10^-31 kg

Assuming the electron starts from rest at the lower plate, its initial kinetic energy is zero.

Step 3: Set the potential energy gained by the electron equal to its final kinetic energy and solve for v.

PE = KE
(1.8x10^-19 C) * (5.25x10^-1 V) = (1/2) * (9.8x10^-31 kg) * v^2

Simplifying the equation:

0.945x10^-19 V = (4.9x10^-31 kg) * v^2

Dividing both sides by 4.9x10^-31 kg:

v^2 = (0.945x10^-19 V)/(4.9x10^-31 kg)
v^2 = 1.93x10^12 m^2/s^2

Taking the square root of both sides:

v = √(1.93x10^12 m^2/s^2)
v ≈ 4.39x10^6 m/s

Therefore, the speed of the electron on reaching the upper plate is approximately 4.39x10^6 m/s.

To solve this problem, we can use the formula for electric potential energy and conservation of energy.

1. First, let's calculate the electric potential energy (PE) of the electron at the lower plate. The formula for electric potential energy is: PE = qV, where q is the charge and V is the potential difference between the plates.

PE = (1.8x10^-19 C)(5.25x10^-1 V) = 9.45x10^-20 J

2. Next, we can find the velocity (v) of the electron as it reaches the upper plate using the conservation of energy. The initial potential energy at the lower plate will be converted into kinetic energy at the upper plate.

PE = KE

KE = (1/2)mv^2

9.45x10^-20 J = (1/2)(9.8x10^-31 kg)v^2

Solving for v^2:

v^2 = (2)(9.45x10^-20 J) / (9.8x10^-31 kg)

v^2 = 1.92x10^11 m^2/s^2

3. Finally, take the square root of both sides to find the velocity (v):

v = √(1.92x10^11 m^2/s^2)
v ≈ 4.38x10^5 m/s

So the speed of the electron on reaching the upper plate is approximately 4.38x10^5 m/s.

Therefore, your answer of 4.3x10^5 m/s is correct. Great job!