Question

Electric field created by an infinite plate that is positively charged is given as 4 N/C. If a negatively charged particle of1.6 C charge and mass 44 Kg held at a distance of 11 meters is let go and moves towards the positively charged plate due to the attractive force, what will be its velocity in m/s when it is 42% of the way on its way to the positively charged plate?

Answer 1

The velocity of the negatively charged particle when it is 42% of the way to the positively charged plate is approximately 6.085 m/s.

Step-by-step explanation:

The electric field created by an infinite positively charged plate is given as 4 N/C. If a negatively charged particle of 1.6 C charge and mass 44 Kg is held at a distance of 11 meters from the plate and then released, it will experience an attractive force and accelerate towards the plate. To find its velocity when it is 42% of the way on its way to the positively charged plate, we can use the equations of motion.

Let's assume the initial velocity of the particle is zero. Since the distance traveled is 42% of 11 meters, it will be 0.42 x 11 = 4.62 meters. Using the equation of motion, v^2 = u^2 + 2as, we can calculate the final velocity:

`v^2`= 0 + 2 × a × s

`v^2 = 2 x 4 N/C X 4.62 meters`

`v^2 = 36.96 N^2/C^2 meters^2`

`v = √36.96 N^2/C^2 meters^2`

`v = 6.085 meters/second`

Answer 2

The velocity of the particle when it has traveled 42% of the distance is 1.18 m/s.

How to calculate the speed of the particle?

The speed of the particle is calculated by applying Coulomb's law of electrostatic force and Newton's second law of motion as follows.

F = EQ

where;

• E is the electric field
• Q is the charge

Also, from Newton's second law:

F = ma

where;

• m is the mass of the particle
• a is the acceleration

ma = EQ

a = EQ/m

a = (4 x 1.6 ) / (44)

a = 0.15 m/s²

The velocity of the particle when it has traveled 42% of the distance is;

s = 0.42 x 11 m

s = 4.62 m

v² = u² + 2as

where;

• u is the initial velocity = 0 m/s
• v is the final velocity
• a is the acceleration
• s is the distance

v² = 0² + 2as

v² = 2as

v = √2as

v = √ (2 x 0.15 x 4.62)

v = 1.18 m/s