Four dipoles, each consisting with dipole moments with |P| = 2.75 times 10^-10 C

Question

Four dipoles, each consisting with dipole moments with |P| = 2.75 times 10^-10 Cm, are located in the xy - plane with their centers 1.0 mm from the origin, as shown. Each dipole has a separation of d = 0.25 mm between the charges. A sphere passes through the dipoles, as shown in the figure. What is the electric flux through the sphere due to these dipoles (epsilon_0 = 8.85times 10^-12 C^2/N. m2) A very large sheet of a conductor carries a uniform charge density of 4.00 pC/mm^2 on its surfaces What is the electric field strength 3.00 mm outside the surface of the conductor What is the electric field 1.0 m from the surface of the conductor (epsilon_0 = 8.85 times 10^-12 C^2/N.m^2) E = sigma/epsilon_0

Explanation / Answer

Written in exercises has some bugs, I solve the problem step by step

Problem 8

We use Gauss' law

= E . dA = qint / o 1

. qintis the charge for internof the surface, this means than external charge give a field zero

Create a Gaussian surface that matches the circle, the radius is

r= 1 cm = 1 10-2 m

besides the field it is equal to all the charges in the center

the field is a positive charge is radial and normal vector to the surface is a radius of the sphere, therefore the E and dA are parallel, thus

E dA = qint/o

for a sphere

dA = 4 r2

The internal charge is

qint= 4 q

the dipole moment

p = q d

where

q is the value of each charge

d is the separation between them

q= p / d

Equation 1 cleared and we substitute

= qint / o = 4q / o

= 4 p/(d o)

= 4 2.75 10-10 / ( 0.25 10-3 8.85 10-12)

= 11/ 2.21 105

= 4.98 105N m²/C

Problem 9

Part a)

For a very large sheet we can assume that is infinite and forget about the problems of the city. By applying Gauss's law to this plane we create a Gaussian surface to be a cylinder with its axis perpendicular to the blade

E dA = qint/o

as the sheet has two sides

E 2A = qint /o

calculate the charge

= qint/ A

E 2A = A/o

2 E = / o

E = / 2o

data

= 4 10-12 c/ mm² (103 mm/1m)2 = 4 10-6 C/m2

E = 4 10-6 /(2 8.85 10-12)

E = 0.226 106 N/C

This field is entirely correct to a point as close to the surface as 3 mm = 3 10-3 m

part b)

In this case there are two possibilities :

* if the surface (L2 ) is large compared to the distance

L >> 1 m

then the above result is correct

* If this is not the case L <= 1 m

must know the size of the sheet for the integral from -L / 2 to L / 2, the reference system is placed in the center of the sheet

I think the former is correct about the lack of data.

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