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Published byLilian Sorrell Modified over 9 years ago

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…and all the pretty variations… F = k q 1 q 2 r2r2 E = k q 1 r2r2 U = k q 1 q 2 r V = k q 1 r

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F = G m 1 m 2 r2r2 Universal Gravitation Constant (the stuff that creates the force) Mass of each object Distance Between Objects G = 6.67 x 10 -11 N∙m 2 /kg 2

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F = k q 1 q 2 r2r2 (the stuff that creates the force) Coulomb’s Constant Charge of each object Distance Between Objects k = 9.0 x 10 9 N∙m 2 /C 2 Charges are measured in units of Coulombs (C)

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Similarities Differences Essentially the same formula Vary directly as the amount of “stuff” Vary inversely as the square of the distance Constants just set units Gravity force is always toward the other mass, electrostatic force can be either direction depending on sign of the charges Positive (+) means they repel Negative (-) means they attract

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1. A negative charge of -3.0x10 -6 C and a positive charge of 4.0x10 -6 C are separated by a distance of 3 mm. What is the force on the negative charge? (magnitude and direction) 2. What is the force on the positive charge? q 1 = -3x10 -6 C q 2 = 4x10 -6 C r = 3x10 -3 m F = (9x10 9 )(-3x10 -6 )(4x10 -6 ) (3x10 -3 ) 2 F neg charge = 1.2x10 4 N toward the positive charge F pos charge = 1.2x10 4 N toward the negative charge

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F = G m 1 m 2 r2r2 Strength at position where second mass would be Units are N/kg, or m/s 2 g = G m 1 r2r2 F m2m2 =

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F = k q 1 q 2 r2r2 Strength at position where second charge would be Units are N/C, or V/m E = k q 1 r2r2 F q2q2 =

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3. What is the electric field at the location of the positive charge due to the negative charge? 4. A positive 6.0x10 -9 C charge experiences a force of 1.8x10 -5 N to the right. What is the electric field at the location of that charge from other charges? q 1 = -3x10 -6 C q 2 = 4x10 -6 C r = 3x10 -3 m E = (9x10 9 )(-3x10 -6 ) (3x10 -3 ) 2 E = 3.0x10 9 N/C from positive to negative charge or E = -1.2x10 4 4x10 -6 E = 3000 N/C to the right E = 1.8x10 -5 6x10 -9 F = 1.8x10 -5 N q 2 = 6x10 -9 C

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The electric field always comes out of a positive charge and into a negative charge: + - + -

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5. What is the direction of the electric field between the two charges of problem 1? Electric Field always goes from positive to negative!

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…Yay! F = k q 1 q 2 r2r2 E = k q 1 r2r2 U = k q 1 q 2 r V = k q 1 r

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F = G m 1 m 2 r2r2 Potential energy of the pair of masses Units are N·m, or J U = F·r = ·r - G m 1 m 2 r (sign only for convention)

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F = k q 1 q 2 r2r2 Potential energy of the pair of charges Units are N·m, or J U = F·r = ·r k q 1 q 2 r

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1. A negative charge of -3.0x10 -6 C and a positive charge of 4.0x10 -6 C are separated by a distance of 3.0 mm. What is the potential energy of the charge pair? 2. Two positive charges of 6.0 μC are 2.0 cm apart. What is the potential energy of the charge pair? q 1 = -3x10 -6 C q 2 = 4x10 -6 C r = 3x10 -3 m U = (9x10 9 )(-3x10 -6 )(4x10 -6 ) 3x10 -3 U = -36 J (toward each other) q 1 = 6x10 -6 C q 2 = 6x10 -6 C r = 2x10 -2 m U = (9x10 9 )(6x10 -6 )(6x10 -6 ) 2x10 -2 U = 16.2 J (away from each other)

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U = k q 1 q 2 r E = k q1k q1 r2r2 Electric Potential based on charge and distance… Units are J/C, or V V = E·r = ·r V = k q 1 r U q2q2 = …or from potential energy k q1k q1 r

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3. What is the electric potential at the location of the positive charge due to the negative charge? 4. Two parallel plates create an electric field between them of 6.0x10 3 V/m. If the plates are 2.0 mm apart, what is the potential difference between the plates? q 1 = -3x10 -6 C q 2 = 4x10 -6 C r = 3x10 -3 m V = (9x10 9 )(-3x10 -6 ) 3x10 -3 V = -9.0x10 6 V or V = -36 4x10 -6 V = 12 V V = E·r = (6x10 3 )(2x10 -3 ) E = 6x10 3 V/m r = 2x10 -3 m

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5. A +4 nC charge and a -4 nC charge are 1.0 m apart. What is the electric field and electric potential at a point half way between them? ELECTRIC FIELD E 1 = (9x10 9 )(4x10 -9 )/.5 2 = 144 E 2 = (9x10 9 )(-4x10 -9 )/.5 2 = -144 E = E 1 + E 2 = 288 N/C → V 1 = (9x10 9 )(4x10 -9 )/.5 = 72 V 2 = (9x10 9 )(-4x10 -9 )/.5 = -72 V = V 1 + V 2 = 0 V ELECTRIC POTENTIAL 0.5m +4 nC + -4 nC -

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The Big Picture F = k q 1 q 2 r2r2 E = k q 1 r2r2 U = k q 1 q 2 r V = k q 1 r Force Electric FieldPotential Energy Potential Difference E = F q2q2 V = U q2q2 U = F·r V = E·r N V J N/C V/m or VECTOR has DIRECTION SCALAR no DIRECTION

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