Question

A thin conducting square plate 1.0 m on the side is given a charge of-2.0 x 10-6 c. A proton is placed 1.0 en above the center of the plate, what is the acceleration of the proton? (Enter the magnitude in m/s.) magnitude direction Select the plate m/s2

Answer 1

The acceleration of the proton is approximately
`-1.08 x 10^13 m/s^2.`

Step-by-step explanation:

To find the acceleration of the proton, we can first calculate the electric field created by the charged plate using the equation

`E = k*q/d^2,`

where E is the electric field, k is the Coulomb constant
`(k = 9 x 10^9 Nm^2/C^2)`, q is the charge of the plate, and d is the distance between the plate and the proton.

Since the plate has a charge of -2.0 x 10^-6 C and the proton is placed 1.0 cm above the center of the plate, the distance between them is 0.01 m. Substituting these values into the equation, we get:

E =
`(9 x 10^9 Nm^2/C^2)(-2.0 x 10^-6 C) / (0.01 m)^2`

Calculating, we find that the electric field is
`-1.8 x 10^6 N/C.`

The acceleration of the proton can then be found using the equation F = m*a, where F is the force on the proton, m is the mass of the proton and a is the acceleration.

The force on the proton can be calculated using the equation F = q*E, where q is the charge of the proton
`(q = 1.6 x 10^-19 C).`

Substituting the values, we get:

`F = (1.6 x 10^-19 C)(-1.8 x 10^6 N/C)`

Then, we can solve for the acceleration:

a = F/m

Plugging in the values, we find that the acceleration of the proton is approximately
`-1.08 x 10^13 m/s^2.`

Answer 2

Acceleration,
`a=1.08* 10^(13)\ m/s^2`

Step-by-step explanation:

It is given that,

Side of the square plate, l = 1 m

Charge on the square plate,
`Q=-2* 10^(-6)\ C`

Position of a proton, x = 1 cm

The electric field due to a parallel plate is given by :

`E=(Q)/(2A\epsilon_o)`

Electric force is given by :

F = q E

`F=(Qe)/(2A\epsilon_o)`

e is the charge on electron

The acceleration of the proton can be calculated as :

`a=(F)/(m)`

m is the mass of proton

`a=(Qe)/(2A\epsilon_o m)`

`a=\frac{2* 10^(-6)* 1.6* 10^(-19)}{2(1)^2* 8.85* 10^(-12)* {1.67* 10^(-27)}}`

`a=1.08* 10^(13)\ m/s^2`

So, the acceleration of the proton is
`1.08* 10^(13)\ m/s^2`. Hence, this is the required solution.