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Field Definition And Coulomb’s Law 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.) Q1Q2 F 2 r r Charge Q1 Charge Q2 The field strength of an electric field is defined by F E Q (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 With a Radial Field The electric field strength take sthe form of Coulombs law. Why? 1 Q E 40 r 2 + 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 !! Test Charge Uniform electric fields Test charge + + + _ _ _ In a uniform field the field strength at any point is given by Remember this is just the force on a unit charge in the field V E d Remember this only applies where the field lines are parallel Together with these relationships: W V Q Q E 40 r 2 The definition of the volt 1 V E d 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 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. So the electric potential at point P P Q 1 coulomb positive charge 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: 1 Q V 40 r Calculations 4.0μC -6.0μC A B Two charges with the values shown are placed along are separated 100mm from A along the line AB by a distance of 100mm. At what distance does the electric potential reach 0V? When the potential along AB reaches zero VA=VB i.e. Now: VA +VB = 0 1 QA VA 40 rA 1 QB VB 40 rB 1 QA 1 QB 0 40 rA 40 rB 1 QA 1 QB 40 rA 40 rB QA QB rA rB 4.0μC V=0 40mm 60mm A rB QB rA QA B 100mm 6 rB 6.0 10 6 6 rA 4.0 10 4 -6.0μC This ratio tells us that V=0 40mm from A Where the numbers are not as straightforward you can continue as follows: 1 2 rA rB 100 As the total distance between the charges is 100mm 4 rA rB 6 2 rA rB 3 2 Now substituting 2 into 1 rB rB 100 3 5 rB 100 3 300 rB 60 5 rA 100 60 40mm 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 ................................................................................................................... . ................................................................................................................................... The diagram shows a positively charged oil drop held at rest between two parallel conducting plates A and B. A Oil drop 2.50 cm 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)