electric field - portland state university · electric field if an electrical charge q o located at...
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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
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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.
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Q Q
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
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
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Q
E
E q
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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
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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
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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
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Does the electric field exist at only one point?
Does the electric field depend on the magnitude or sign of the test charge qo?
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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?
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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
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ELECTRIC FIELD (Continuation and Review)
e l e c t r i c f i e l d
+ q
E
+ q
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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
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Example: Electric field produced by a positive point-charge q
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Example: Electric filed produced by a negative point-charge Q
-
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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
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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 .
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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
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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
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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
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P
T
S
W
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Charges inside an electric field
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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?
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Electric-dipole inside an electric field
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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
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Electrical Ground: A very large conductor able to supply an unlimited amount of charge Review on Dielectric Breakdown
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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?
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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
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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
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Example 2. Electric field established by auniformly charged disk.
P
A=πR2
Question: What is the direction of the electric field atpoint P?
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dA = 2πr dr
Ring of radius r and thickness dr How much charge (dq) does this ring contains?
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σR2
-2 -2
z
E(z) = R
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