UNIT 2c LO7ENERGY TO LO8 LAPLACE AND POISSON.ppt - LO 2.7 ENERGY IN ELECTROSTATIC FIELDS WORK POTENTIAL ENERGY LO 2.7 Identify Energy in Electrostatic

UNIT 2c LO7ENERGY TO LO8 LAPLACE AND POISSON.ppt - LO 2.7...

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ENERGY IN ELECTROSTATIC FIELDS LO 2.7
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WORK, POTENTIAL & ENERGY LO 2.7 Identify Energy in Electrostatic Fields. Explain the concept of energy exchanges in the electric field of a system of charges. Explain the concept energy expended to build up a system of charges. Perform calculation related to energy in electrostatic fields.
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WORK DONE IN MOVING A POINT CHARGE
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WORK DONE IN MOVING A POINT CHARGE. A charge Q experience a force F in an electric field E. In order to maintain the charge in equilibrium a force F a must be applied in opposition. F a=-Q E F = Q E E F = F a F = Q E F a = -Q E
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Work = a force acting over a distance. Therefore, a differential amount of work, dW is done when the applied force F a produces a differential displacement d of the charge i.e move the charge through the distance. WORK DONE d = |d | dW = F a . d = -Q E . d If Q is +ve & d is in the direction of E , dW = -QE dℓ <o, indicating that work was done by the electric field.
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Component forms of the differential displacement vectors are as follows: d = dx a x + dy a y + dz a z (cartesian) d = dr a r + rdø a ø + dz a z (cylindrical) d = dr a r + rdθ a θ + rsinθdø a ø (spherical) Differential displacement vectors, d Recall
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Cartesian coordinate system Component form of the differential displacement vector, d = dx a x + dy a y + dz a z Component form of the differential volume vector, d v = dxdydz Recall
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Cylindrical coordinate system d = dr a r + rdø a ø + dz a z differential displacement vector d ℓ, d v = rdrdødz differential volume vector d v , Recall
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Spherical coordinate system d = dr a r + rdθ a θ + rsinθdø a ø differential displacement vector d ℓ, d v = r 2 sinθdrdθdø differential volume vector d v , Recall
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An electrostatic field is given by E = (x/2 + 2y) a x + 2x a y (V/m). Find the work done in moving a point charge Q = -20 μC a)From the origin to (4,0,0) m, and b)From (4,0,0) m to (4,2,0) m. WORK DONE EXAMPLE 1
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a) The first path is along the x axis i.e from ( 0 ,0,0)m to ( 4 ,0,0) m. So that, differential displacement, d = dx a x dW = -Q E. d W = -Q ∫ E. d = -(-20 μ)∫{(x/2 + 2y) a x } . {dx a x } = (20 μ)∫(x/2 + 2y)dx = 80 μJ WORK DONE SOLUTION
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b) The 2 nd path is in the a y direction i.e from (4, 0 ,0)m to (4, 2 ,0) m. So that, differential displacement, d = dy a y dW = -Q E. d W = -Q ∫ E. d = -(-20 μ)∫{(2x) a y } . {dy a y } = (20 μ)∫2xdy = 320 μJ WORK DONE SOLUTION
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WORK DONE - Conservation property of the electrostatic field The work done in moving a point dharge from one location, B to another, A, in a static electric field is independent of the path taken. Thus, in term of figure below, 1 2 A B W 1 = W 2 1 E. d ℓ = -∫ 2 E. d Or 1-2 E. d = 0 Where the last integral is over the closed contour formed by 1 described positively and 2 described negatively.
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WORK DONE - Conservation property of the electrostatic field Conversely, if a vector field F has the property that F. d ℓ = 0 over every closed contour, then the value of any line integral of F is determined solely by the endpoints of the path.
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  • Fall '19
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