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Field Definition And Coulomb’s Law

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Presentation on theme: "Field Definition And Coulomb’s Law"— Presentation transcript:

1 Field Definition And Coulomb’s Law

2 Coulomb’s Law Gives us the rule for dealing with two point charges.
(in practice for two charges whose separation is much greater than the radius of the charges.) r Charge Q1 Charge Q2

3 The field strength of an electric field is defined by
(Unit NC-1) That is the force exerted by the field on unit charge (ie a charge of 1 Coulomb) placed at that point. +1C F

4 With a Radial Field Test Charge
The electric field strength take sthe form of Coulombs law. Why? + Q The two variable quantities are the charge Q and the distance r The electric field strength formula is just coulombs law applied to a test charge of 1C !!

5 Uniform electric fields
Test charge _ _ _ In a uniform field the field strength at any point is given by Remember this only applies where the field lines are parallel Remember this is just the force on a unit charge in the field

6 Together with these relationships:
The definition of the volt The electric field strength due to a point charge The electric field strength in a uniform field So the unit of E is Vm-1 as well as NC-1 Form the basis of any solution to the electric fields questions you will be asked

7 Electric Potential The electrical potential of any point in the field is the work done to bring a (+) charge of 1 coulomb from infinity (i.e. beyond the influence of the field) to that point in the field. Q So the electric potential at point P 1 coulomb positive charge P

8 Implications: 1.The electrical potential of any point beyond the field is zero 2. The electric potential is the potential energy change for 1C of charge The electric potentila is given by:

9 Calculations 4.0μC -6.0μC A B Two charges with the values shown are placed along are separated by a distance of 100mm. At what distance from A along the line AB does the electric potential reach 0V? 100mm When the potential along AB reaches zero VA=VB i.e VA +VB = 0 Now:

10 V=0 4.0μC 40mm 60mm -6.0μC A B 100mm This ratio tells us that V=0 40mm from A

11 Where the numbers are not as straightforward you can continue as follows:
As the total distance between the charges is 100mm 1 2 Now substituting 2 into 1

12 Calculate the magnitude of the electric field strength at the surface of a nucleus U (Z=92 M=238) . Assume that the radius of this nucleus is 7.4 × 10–15 m. Magnitude of electric field strength = State the direction of this electric field. State one similarity and one difference between the electric field and the gravitational field produced by the nucleus. Similarity Difference .

13 ………………………………………………………………………………………………
The diagram shows a positively charged oil drop held at rest between two parallel conducting plates A and B. The oil drop has a mass 9.79 x 10–15 kg. The potential difference between the plates is 5000 V and plate B is at a potential of 0 V. Is plate A positive or negative? ……………………………………………………………………………………………… Calculate the electric field strength between the plates. Electric field strength =………………………………… Calculate the magnitude of the charge Q on the oil drop. Charge =…………………………………… How many electrons would have to be removed from a neutral oil drop for it to acquire this charge? (3)


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