Two large parallel non-conducting sheets are spaced d = 2.40 cm apart and have a

Question

Two large parallel non-conducting sheets are spaced d = 2.40 cm apart and have a potential difference of magnitude 625 V. An electron is thrown directly away from one sheet, and is stopped just before reaching the other sheet. What is the initial speed given to the electron? The electron must have been thrown: What is the strength of the electric field between the plates? If you are told the plates have equal but opposite charge density, what must be its magnitude? However, the two plates could also be charged in which of the following ways and still cause this potential difference? Make a sketch of the electric field vs. x and voltage vs. x. (Take the +x axis to intersect both sheets perpendicularly, running in the same direction as the electron's initial velocity.) What is the shape of the E vs, x graph, for x values that are: outside the plates: between the plates: What is the shape* of the V vs. x graph, for x values that are outside the plates: between the plates: Suppose you instead had two other large sheets like this, with the SAME potential difference between them, but with double the separation. If you again wanted to throw an electron from one sheet and have it be stopped just before reaching the other, what - would your answers (as in parts a-c) be now? (Try to conceptualize if/why your answers increase or decrease.)

Explanation / Answer

A) For this problem we will use the second law of Newton, where the force is electric

F= m a

- e E = m a

a = -e E / m

We use expressions kinematics

Vf2= Vo2+ 2ax

As the electron stops Vf =0

0 = Vo2+ 2ax

Vo2= -2 a x

Vo2= -2 (-eE/m) x

We calculate the electric field from the potential

V= - E x

E = V /x

Vo2 = 2 e (V/x) x/m

Vo2= 2 e V /m

Vo2 = 2 1.6 10-19 625 / 9.1 10-31

Vo2= 219.78 1012

Vo = 14.82 106m/s

The electron moves from the negative plate to the positive field is field or from low potential move to higher voltage plate

B) The expression of the electric field

V= 0-625 = -625 V

d = 2.40 cm = 2.40 10-2 m

V = - E d

E = - V /d

E = 625 / 2.40 10-2

E =26041.6 N/C

C) For this part we can use the law of electric field Gauss

= E A = Qint /o

We assume that the plates are very large compared with the dimensions of the exercise

Create a Gaussian surface is a cylinder with parallel faces to the plate

2E A = Qint /o

the two must electric field to both sides of the plate.

The surface charge density is

= Qint / A

calculate

2E A = A /o

2E = /o

= 2 E o

= 2 26041.6 8.9 10-12

= 4.63 10-7C/m²

We can have various forms of load plates when charged with charges of equal density but opposite direction each generating one half the electric field which introduces another factor ½

I mean that the charge density is total, so each plate has half of this value

(RESULT)

(plate) = 2.365 10-7C/m

Another way to charge plates Cero Positive

D)

Between the plates. The electric field lines are straight from the positive plate to the negative

Outside the plates .. The electric field lines are emerging from the positive plate and give a curve to reach the negative plate

between the plates. They are equipotential surfaces are parallel flat plates, the lower potential near the negative plate

Outside the plates are not flat and deformed to maintain the potential and are perpendicular to the field lines.

e) X = 2 X0 =

Vo = 14.82 106 m/s

E = - V /d

E = 625 / 4.80 10-2

E =13020.8 N/C

= 2 E o

= 2 13020.8 8.9 10-12

= 4.63 10-7C/m²

(plate) = 1.16 10-7 C/m²

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