Charge in a regular electric field

A regular electric field has a value of 5000 N/C. A particle with mass 2,5 x 10^-6 kg and charge
 1 x 10 ^-8 C is placed in the point A of the field. If the particle is free, it will move towards a point B.
a) Calculate the acceleration of the particle.
b) Calculate its speed for 1 μs.


E= 5000 N/C= 5 ∙ 10 ³  N/C
m= 2,5 x 10^-26 kg

q=  1 x 10 ^-8 C

a) a=?

b) v=? if t=1 μs= 10^-6  s

SOLUTION

a) According to Newton's Second Law of Motion:  F=m ∙ a <=> a= F/m = q ∙ E/m

=> a= 10 ^-8  ∙ 5 ∙ 10 ³ / 2,5 x 10^-26  = 2 ∙ 10 ²¹ m/s²

b) v= at= 2 ∙ 10 ²¹∙  10^-6= 2 ∙  10¹⁵ m/s

Advanced Physics Problem

Three spheres with radius r, 2r and 3r, have charges 3q, -2q and 3q reciprocally. They are placed in the angles  of a right quadrilateral  pyramid , with radius R>>>r.

Calculate the electrical field  in the fourth angle of  the  right quadrilateral pyramid.

r 

2r

3r

3q

-2q

3q

R>>>r

E_A=?

The condition R>>>r allows us to neglect the distribution of charges in the spheres and consider that they are uniformly spread in the spheres’ surfaces. It can be deduced that moduls of intensities E₁ and E₃  are equals, because the charges and the distances which cause them in point A are equals.  

According to the superposition principle of the electric fields, we have:

The modules of vectors  and    are equals and the angle between them is 60 degrees. Their resultant vector   is the rhombus’ diagonal which has these vectors as ribs, as a result its module is:  E _{1 3} = 2∙ E₁∙cos30


In order to calculate  E_{A ,} we should add vectorialy  with  .  These vectors lie in the plane ABF (AF is the height in the equilateral triange ACD).

According to cosinus theorem:
E_A= E_{1 3} ² + E₂ ² - 2 E _{1 3}∙E₂ cos β

Angle β as we can see in the image is equal to the angle between the rib and the pyramid’s  side.

Looking the triangle ABF and knowing that AF=BF,

cos β= 1/2cos 30

Using this equation and the one for E_{1 3} , after easy conversions, we can find
 E_A= 3 E₁ ² + E₂ ² - 2E₁E₂ (1)
 E₁=E₃= k 3q/R² ; E₂=k 2q/ R² (2)
 By substituting in equation 1 the equations (2), we find the electric field in point A: E_A= k q 19/R²

E_A= 19k q/R²

How to use integration by parts method?

How to use integration by parts method?

General rule: ∫udv=uv-∫vdu

When we integrate by parts ∫f(x)dx , we express the f(x)dx as the product of the factors u and v.

Then we make 2 integrations:

1) From the differential dv we find v.

2) We find ∫vdu

We use this method when finding the above integrals is simpler then finding the integral of the whole expression.

There are no strict rules which indicate what part of the product we will substitute with u and what with v. However, there are some typical integrals for which we can define this, such as:

a) ∫x cosbxdx

b) ∫xsinbxdx

c) ∫x e^ax dx
d) ∫x^a  lnx dx;
For a,b,c we make the substitutions: u=x and dv=cosbxdx, dx=sinbxdx, dx=e^ax dx. However, in c) we substitute lnx=u and dv=x^a  dx.

Examples:

1. ∫lnxdx
u=lnx
dv=dx <=> v=∫dx=x
Integration by parts rule: ∫udv=uv-∫vdu
∫lnxdx= xlnx-∫xd(lnx)  d(lnx)= lnx'dx=dx/x
∫lnxdx= xlnx-∫x dx/x = xlnx - ∫dx= xlnx -x +c

2. ∫(x+1)sinx dx
x+1=u
dv=sinxdx <=> v=∫sinxdx= -cosx
∫(x+1)sinx dx= -(x+1)cosx - ∫-cosxdx= -(x+1)cosx - (-sinx) + c= -(x+1)cosx + sinx + c

Interference exercise 5

Two holes placed at a 2 x 10^-4 m distance from each other are lit by a monocromatic light. The direction to the second minimum forms a 0.254degrees angle relative to the perpendicular to the screen. Calculate the wavelength

Solution

d=2 x 10^-4m
α=0.254 degrees
k=1
λ=? Minimums interference formula:
dxsinα_k=(2k+1)λ/2 <=> dxsinα=3λ/2 <=> λ=2 x d x sinα/3 <=>λ=2 x 2 x 10^-4 x sin(0.254)/3= 5.91 x 10^-7m= 591 nm

Interference exercise 4

Two Young interference holes produce the first maximums in angles +-12degrees relative to the perpendicular line to the screen. The light wavelength is 546nm. Calculate the distance between the holes.

Solution

λ=546nm=5.46 x 10^-7m
α=12degrees
d=?
Formula for calculating interference maximums: dxsinα_k=kλ (k=1)
d=λ/sinα=5.46 x 10^-7/0.21= 2.6 x 10^-6m

Interference exercise 3

Two holes placed at a 0.8 mm distance from each other, are lit by a monocromatic light with wavelength 720nm. The distance between two interference ribbons is 4mm. Calculate the distance L between the screen and the holes.

Solution

d=0.8 mm= 8x10^-4m
λ=720nm=7.2x10^-7m
The distance between two lit ribbons is equal to the distance between the first and the central lit ribbon, y₁=4mm=4x10^-3m
Lit interference ribbons formula:
dxsina_k=kλ, dxsina₁=λ
for small angles sina=tga=y1/L
dxy₁/L=λ <=> L=dxy₁/λ=8 x 10^-4 x 4 x 10^-3/7.2 x 10^-7=4.44m

Interference exercise 2

2. A monocromatic light produces an interference pattern on a screen placed at a 2.5 m distance from the holes, which have a 0.8mm distance from each other. The y1 distance between the first lit ribbon and the central one is 2mm. Calculate the wavelength.

Solution

λ=?
y₁=2mm= 2 x 10^-3m
L=2.5m
d=8 x 10 ^-4m
Formation of lit interference ribbons formula:
d x sinα_k=kλ, for small angles: dxtgα_k=kλ (tgα_k=y_k/L)
λ=dxtgα_k/L
λ= 8 x 10 ^-4x2x10^-3/2.5= 6.4x 10^-7m