An electron in a computer monitor enters midway between two parallel oppositely charged plates. The initial speed of the electron is 6.15 x 10^7 m/s and its vertical deflection is 4.70 mm. (a) What is the magnitude of the electric field between the plates? (b) Determine the magnitude of the surface charge density on the plates in C/m^2. The length of the plates are 10 cm and the width of the plates are 1 cm.

To find the magnitude of the electric field between the plates, we can use the equation for the vertical deflection of a charged particle in an electric field:

d = (1/2) * (e/m) * E * t^2

Where:
d = vertical deflection (4.70 mm = 0.00470 m)
e/m = charge-to-mass ratio of the electron (-1.76 x 10^11 C/kg)
E = electric field (what we want to find)
t = time of flight inside the plates

First, let's calculate the time of flight (t) using the initial speed of the electron:

t = (2 * d / E)^0.5

Now, substitute the known values into the equation to find t.

t = (2 * 0.00470 m / E)^0.5

Next, let's calculate the value of t.

t = (2 * 0.00470 m / E)^0.5
t^2 = (2 * 0.00470 m / E)
t^2 = 2 * 0.00470 m / E
E * t^2 = 2 * 0.00470 m
E = 2 * 0.00470 m / t^2

Now, substitute the known values into the equation to find E.

E = 2 * 0.00470 m / t^2
E = 2 * 0.00470 m / (6.15 x 10^7 m/s)^2

Calculating the squared value of t:

t^2 = (6.15 x 10^7 m/s)^2
t^2 = 3.77225 x 10^15 m^2/s^2

Substituting the value of t^2 back into the equation:

E = 2 * 0.00470 m / (3.77225 x 10^15 m^2/s^2)
E = 9.82703 x 10^-16 N/C

Therefore, the magnitude of the electric field between the plates is 9.82703 x 10^-16 N/C.

Now, let's move on to calculating the magnitude of the surface charge density on the plates.

Surface charge density is defined as the charge per unit area. We can express it using the formula:

σ = Q / A

Where:
σ = surface charge density (what we want to find)
Q = total charge on either plate
A = area of either plate

The total charge on either plate can be calculated as the product of the electron charge and the number of electrons hitting the plate.

Q = (e/m) * (electron speed) * (time of flight)

Substituting the known values:

Q = (-1.76 x 10^11 C/kg) * (6.15 x 10^7 m/s) * t

Now, let's substitute the value of t calculated earlier:

Q = (-1.76 x 10^11 C/kg) * (6.15 x 10^7 m/s) * [(2 * 0.00470 m) / E]^0.5

Calculating Q:

Q = (-1.76 x 10^11 C/kg) * (6.15 x 10^7 m/s) * [(2 * 0.00470 m) / (9.82703 x 10^-16 N/C)]^0.5

Next, let's calculate the area of either plate:

A = length * width = (0.10 m) * (0.01 m)

Now, substitute the known values into the equation to find σ.

σ = Q / A
σ = {(-1.76 x 10^11 C/kg) * (6.15 x 10^7 m/s) * [(2 * 0.00470 m) / (9.82703 x 10^-16 N/C)]^0.5} / (0.10 m * 0.01 m)

Calculating the value of σ:

σ = (-1.76 x 10^11 C/kg) * (6.15 x 10^7 m/s) * [(2 * 0.00470 m) / (9.82703 x 10^-16 N/C)]^0.5 / (0.10 m * 0.01 m)

Therefore, the magnitude of the surface charge density on the plates is the calculated value of σ.

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