PHY122_L03 - PHY122 Physics for the Life Sciences II PHY122...

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HY122 HY122 - hysics for the Life Sciences II PHY122 PHY122 Physics for the Life Sciences II Physics for the Life Sciences II Lecture 3 The Electric Potential Lecture 3 1 09/06/2011
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Channel Setting Instructions for ResponseCard ResponseCardNXT NXT 1. Press and release the “ Channe l" button. 2. Use the number pad to enter the 2 digit channel code (i.e. channel 21 = 21): ur Channel is 21 Our Channel is 21 nce the channel number has been typed press 3. Once the channel number has been typed, press the (c)enter button. The should appear to confirm success. 09/06/2011 Lecture 3 2
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Two point charges repel each other with a 2.0 force N force. Then, I decrease the separation by a factor 3 he repulsive force now becomes N? The repulsive force now becomes … N? 18 59 62 0 26 96 THER 7 OTHER 27 Matched Acceptable Value: 59, Within Range: 59 cceptable Value: 18 Acceptable Range: {17 9 18 1} 09/06/2011 Lecture 3 3 Acceptable Value: 18, Acceptable Range: {17.9, 18.1}
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Coulomb Force he oulomb orce is the electrostatic (electric) force The Coulomb force is the electrostatic (electric) force between two point -like charges Q 1 and Q 2 , separated by a center-to-center distance r 2 . 12 The Force magnitude on Q 2 from Q 1 : irection: 12 2 C QQ r FK Coulomb’s Law Direction: Repulsive for LIKE = same-sign charges Attractive for UNlike (opposite-sign) charges – Coulomb constant: permittivity of vacuum: ote that this is imilar to gravity etween on masses rom 92 2 0 1 8.998 10 Nm /C 4 K   2 2 0 8.85 10 C /(Nm )  Note that this is similar to gravity between on masses M 2 from M 1 : 2 G M M r FG Newton’s Law of Gravity Lecture 3 4 •n o t e that the masses act like gravitational “charges”! • (BUT: only positive mass exists and gravity is always attractive!) 09/06/2011
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Electric Electric Field Field E ( r ) For complex configurations of charges which are fixed, it is useful to use the ELECTRIC FIELD as intermediary – i.e. the force-per-unit-test-chargeeverywhere in space: 2 2 ) () ( i qi i qi i qi i qi Q q q Q K r K r  Fr r Er r Once we have calculated the FIELD E ( r ) from charges Q i we can find the force on ANY test-charge q test , NYWHERE simply as = ANYWHERE, simply as F ( r , q test ) = q test E ( r ) a ECTOR FIELD E ( r ) is a VECTOR FIELD : E = ( E x ( r ), E y ( r ), E z ( r ) ) ; with r = ( x,y,z ) Lecture 3 5 09/06/2011
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