PHY132_L03 - Classical Physics II Classical Physics II...

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lassical Physics II Classical Physics II PHY132 Lecture 3 The Electric Field Lecture 3 1 01/29/2010
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Recap: Coulomb Force he Coulomb force is the electrostatic (electric) force • The Coulomb force is the electrostatic (electric) force between two point charges Q 1 and Q 2 , separated by a center-to-center distance r 12 , which points from Q 1 to Q 2 . The Force from Q 1 on Q 2 is then: l 12 2 1 2 1 (Coulomb's Law) 4 C QQ r πε = Fr ote that this is milar to gravity etween two masses nd 01 92 2 1 2 2 2 0 0 1 8.998 10 Nm /C ; 8.85 10 C /(Nm ) 4 ε = × – Note that this is similar to gravity between two masses M 1 and M 2 : ote that the masses are the gravitational “charges”! (BUT: only ositive l 2 G MM G r =− • note that the masses are the gravitational charges ! (BUT: only positive mass exists) – But the electrical force between two protons is MUCH STRONGER an the gravitational force: Lecture 3 2 than the gravitational force: 36 0 11 2 4 9.0 10 1.2 10 6.67 10 C Gp FQ Q e FG M M m × == = × × 01/29/2010
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Recap: Electric Recap: Electric Field Field E ( r ) • When we have even more complex configurations of charges which are fixed, and if their effect on a “test- charge” is to be calculated, it is often easier to “pre- calculate” the ELECTRIC FIELD , i.e. the force-per-unit- test-charge everywhere once and for all; nc h v c lcul t d th FIELD •once we have calculated the FIELD E ( r ) F ( r , q test )/ q test , we can find the force on ANY test- charge q test , ANYWHERE, simply as F ( r , q test ) = q test E ( r ) •Thus : E ( r ) is a VECTOR FIELD : E = ( E x ( r ), E y ( r ), E z ( r )); with r =( x,y,z ) • A similar situation exists for gravity; e.g. near sealevel: E ( r ) F ( r , m st )/ m st = – m st g j /m st = – g j Lecture 3 3 G G test test test test the gravitational force is simply found by multiplying by the “gravitational charge” m test : F G = m test E G = – m test g j 01/29/2010
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The Electric Field of a Uniform Line Charge • Field as function of distance y to line charge: – This makes sense: im si s OK: [ is / s h th s l / t h s d 0 1 2 y E y λ πε = • Dimensions OK: λ ] is C/ m , so we have the usual C/m 2 for the second factor … • The distance behavior is 1/m , VERY different from the 1/m 2 for a point harge! charge! • This also makes sense: the SYMMETRY of the problem is cylindrical, and the field lines can only spread out perpendicular to the line; • Thus the field line DENSITY decreases with distance as 1/circumference of a circle (NOT as 1/surface of a sphere!) • Thus, the behavior MUST be like 1/distance from the line (NOT like 1/distance-squared!) • We’ll now look at the field of a uniform circular line of
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This note was uploaded on 04/14/2010 for the course PHY 132 taught by Professor Rijssenbeek during the Spring '04 term at SUNY Stony Brook.

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PHY132_L03 - Classical Physics II Classical Physics II...

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