P

Operational procedure to calculate the ELECTRIC FIELD produced a given system of charges at the point P

system ofcharges

i) Place a test charge qo at the point P. ii) Find the electrical force F that such system of charges exerts on qo. iii) The ELECTRIC FIELD at P will be given by

P P

The ELECTRIC FIELD is a vector

qo E

Andres La Rosa Lecture Notes Portland State University PH-212

ELECTRIC FIELD If an electrical charge qo located at a position P experiences a force, we say that there exist an electric field in a region around P. The electric field is produced by the system of charges. The electric field characterizes the system of charges.

qo

= E

First, some basic concepts about vectors

Next, we show that the electric field produced by a positive charge Q is independent of the sign of the test charge used to evaluate it.

Q Q

QQ

Now we show that the electric field produced by a negative charge q is independent of the sign of the test charge used to evaluate it.

q

qq

qo (+)

q1 (-)

Positive test charge qo

Negative test charge q1

E

E

Positive test charge q3

Negative test charge q4

q3 (+)

q

E

E

F

F

F Force

Force

q4 (-)

PP

P

P

Q

E

E q

�

First, let's place a positive test charge qo at the position indicated by P, and evaluate the total electric force acting on that charge.

Example. Indicate the electric filed at the point P established by the two positive charges shown in the figure

P

qo

Notice, there is no charge at P

F

Next, we simply calculate the ratio ( Since qo is positive, then E and F will be oriented in the same direction)

E

Force vector

Electric field vector

P

PNotice, there is no charge at P

E qo F

E

�

Had we have chosen a negative test charge qo, we would have obtained the following

( Since qo is negative, then E and F will be oriented in the opposite direction)

qo FForce vector

EP

Conclusion E does not depend on the "test" charge qo.

System of charges

E

E

Exercise. Find the electric field at point P(0,4) and at the point T(0,-2)

q1 q2X (cm)

Y (cm)

System of charges

Solution E01 = 9x109 (10x10-6) / (5x10-2)2 = 3.6 x 107 N/C Unit vector u1 = (3/5, 4/5) = ( COS 530, SIN 530 ) E02 = 3.6 x 107 N/C Unit vector u2 = (3/5, - 4/5) E (at P) = 2 x 3.6 x 107 N/C x (3/5, -0 ) = (4.32 x 107 N/C , 0 )

q1 = 10 μC q2 = - 10 μC

Does the electric field exist at only one point?

Does the electric field depend on the magnitude or sign of the test charge qo?

check

System of charges (Charges on a stick bar)

How can we mathematically verify that the electric field of a system of charges is independent of the test charge qo alluded in the experimental procedure given at the beginning of this chapter?

Summary

System of charges

E

E

E

If a charge q (positive or negative) were placed at the position P shown in the figure above (and provided that such charge does not disturb the position of the system of charges) then q will experience a force given by,

q EEF =

P

�

ELECTRIC FIELD (Continuation and Review)

e l e c t r i c f i e l d

+ q

E

+ q

How to calculate the electric field E at a point P located a distance "r" from a positive point-charge of magnitude e ?

+ e

Units e :

How to calculate the electric field E at a point S located a distance "r " from a negative point-charge of magnitude e ?

- e

E (at S) =

E

E (at P) = E E E

S

rr

Example: Electric field produced by a positive point-charge q

Example: Electric filed produced by a negative point-charge Q

-

What is a DIPOLE ?

- Q Charge induction inside the molecule results in the creation of a dipole. The dipole orients in response to the external electric field

Charge-neutral molecule

Microscopic description of non-conductivity, in terms of the electric-dipole concept

- +

Definition of the electric dipole moment:

is a vector that point from the negative charge towards the positive charge

non-conducting material

The dielectric breakdownThere exists a maximum value for the external electric field beyond which the atoms inside the material become ionized. The dislodged electrons (now free) give rise to a high current, which heats up, and eventually destroy, the material. This phenomenom is called dielctric breakdown.

Dielectric breakdown occurs in air at E = 3x 106 N/C .

The water molecule

H

H

Oxygen

Negative side

Positive side

Accordingly, the molecule of water can be treated as an electric dipole

p = 6.2 x 10-30 Cm

For a neutral water molecule in its vapor state

Real structure

Simplified model

d = 0.04 Angstroms

d =

d

d

10 protons (+) 10 electrons (-)

p = Q d = 6.2 x 10-30 Cm = (10e) d

Electric field established by a dipole

Electric field at different locations (P, T , S) produced by the charges +Q and - Q

+Q

- Q

P

T

S

E

E

E

P

T

S

W

Charges inside an electric field

Charge moving inside a uniform electric field

E EX

Y

E = 1.4 x 106 N/C Negative charge of mass m

Question: What would be trajectory followed by the charged particle? Does the acceleration of the particle change during its the motion inside the field?

Calculate: i) The acceleration of the particle ii) How much does the particle deflects (vertically) when it travels a horizontal distance of 1.6 cm from the instant it enters the region of the electric field?

�

Electric-dipole inside an electric field

Order of magnitude of the electric force

Repulsive force between 2 protons inside a nucleus

Attractive force between a proton in the nucleus of an atom and an electron flying around the nucleus

Electric field established by a proton at a distance 1 Angstrom far away

Electrical Ground: A very large conductor able to supply an unlimited amount of charge Review on Dielectric Breakdown

Exploiting symmetry (of the charge distribution in a system of charges) to calculate electric fields

Example 1. Electric field established by auniformly charged circular line.

P

A total amount of positivecharge Q is UNIFORMLYdistributed along a plasticring of radius R

Question: What is the direction of the electric field atpoint P?

First

Second

Third: Apply symmetry

For each small segment containing a chargedq1, there exists another segmentsymmetrically located containing a chargedq2 (with dq2=dq1)such that the HORIZONTALcomponents of the electric field cancel out.That is,only the VERTICAL components makea net contribution to the TOTAL FIELD.

Solution

Thus, based on the grounds of symmetry,we calculate the VERTICAL component ofthe electric field produced by a charge dqcontained in a segment of length ds, andthen ADD UP all those verticalcomponents correspondent to eachcharge on the circular line

Example 2. Electric field established by auniformly charged disk.

P

A=πR2

Question: What is the direction of the electric field atpoint P?

dA = 2πr dr

Ring of radius r and thickness dr How much charge (dq) does this ring contains?

σR2

-2 -2

z

E(z) = R