PHY122_L02

PHY122_L02 - 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 2 The Electric Field Lecture2 1 09/01/2011

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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 hannel is 1 Our Channel is 21 nce e channel number has been ped ress 3. Once the channel number has been typed, press the (c)enter button. The should appear to confirm success. 09/01/2011 Lecture2 2
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 , which points 12 from Q 1 towards Q 2 . The Force on Q 2 from Q 1 : epulsive or IKE ame gn charges 12 2 2 1 C QQ r K Fr Coulomb’s Law – Repulsive for LIKE = same-sign charges – Attractive for UNlike (opposite-sign) charges – Coulomb constant: 2 1 92 2 1 8.998 10 Nm /C K  2 2 2 permittivity of vacuum: 0 4  0 8.85 10 C /(Nm ) – Note that this is similar to gravity between on masses M 2 from M 1 : 1 2 2 G MM r G  Newton’s Law of Gravity Lecture2 4 • note that the masses act like gravitational “charges”! • (BUT: only positive mass exists and gravity is always attractive!) 09/01/2011

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Example Consider two charges Q 1 = +5 μ C , Q 2 = –8 μ C , 5 cm apart. alculate the force on a third charge = 1 cm rom Calculate the force on a third charge Q 3 +1 μ C , 3 cm from Q 1 , 4 cm from Q 2 : F 1 = 9×10 9 ×(5×10 6 (1×10 6 ) / 9×10 4 = 50 N F 2 = 9×10 9 ×(8×10 6 (1×10 6 ) /16×10 4 = 45 N F total, x = F 1 x + F 2 x = 50 cos θ 1 + 45 cos θ 2 F 1 magnitude!! = 50 (3/5) + 45 (4/5) = 66 N i F total, y = F 1 y + F 2 y = 50 sin θ – 45 sin θ y F 2 F tot 1 2 = 50 (4/5) – 45 (3/5) = 13 N j r graphically: 4 cm 3 cm x Lecture2 6 or graphically: 09/01/2011 Q 1 =+5 μ C Q 2 = –8 μ C 5 cm
Example Consider two charges Q 1 = +5 μ C , Q 2 = –8 μ C , 5 cm apart. • Question: where can I place a third charge q so that is eels zero net force? feels zero net force? Q 1 =+5 μ C Q 2 = –8 μ C 5 cm Solution: • not in-between … 58 ? 0 CC   FF 22 ? 5/ (8 )/ ( 5 ) 0 Kqx K q x  Note: use units μ C and cm ! • not to the right of Q 2 • to the left of Q 1 ? ? 8/( 5) 0 xx  ? 5( 8 0 • Answer: 1.17 cm to the left of Q 1 ?

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PHY122_L02 - PHY122 Physics for the Life Sciences II PHY122...

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