# 325_Sp2011_4_Potential_and_gradient - Electric Field Flux...

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© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin Electric Field Flux, Gauss’s Law, Divergence 1 4. Electric potential and gradient. Electric dipole. EE325 Mikhail Belkin

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© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin Electric Field Flux, Gauss’s Law, Divergence 2 Work done in a force field work = force · distance The work is negative if force is applied in the direction opposite to travel direction Work done by an external source increases (decreases if work is negative) the energy of an object that we moved What if the force varies along the travel path? We have to do integration. Work done depends only on the component of force along the distance traveled ( F is the force that we apply to move an object by a vector ) The total work done is a line integral along the path To move a charge Q in electric field, we need to apply force F =-Q E (we work against the E-field) path W Q E dl = - r r dW F dl = r r = l d F W r r l d r To better understand the concept, imagine we are moving a positive charge in towards another positive point charge at the origin. We work against E-field and increase the energy of a system of charges (they now repulse more and will fly away faster when released). Similarly, we if more the charge away from the charge at origin, we will do a negative work and decrease the energy of a system of charges.
© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin Electric Field Flux, Gauss’s Law, Divergence 3 The line integral initial position

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© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin Electric Field Flux, Gauss’s Law, Divergence 4 example: moving a charge in a line-charge field path: constant ρ , around an arc of a circle obviously the work is zero Similarly, work is also zero when path is vertical path: constant φ , from ρ = b to ρ = a ( 29 1 ˆ ˆ 2 a L b o path W Q E dl Q d ρ ρ ρ ρ πε ρ = - = - ÷ r r 1 ln 2 2 a L L b o o a Q d Q b ρ ρ ρ πε ρ πε = - = - z y x a ˆ z ˆ ρ ˆ φ ˆ E E ρ ρ = r 1 2 ˆ L line charge o E ρ ρ πε ρ = r b check the sign! we moved charge from b to a ! integration order gets path direction ρ ˆ along directed is l d r
© Copyright Dean P. Neikirk 2004-2009 Mikhail Belkin, EE 325, ECE Dept., UT Austin Electric Field Flux, Gauss’s Law, Divergence 5 Conservative fields and potential difference If the work done for moving an object between in a force field any two points A and B is independent of the path taken the force field is “conservative” The work integral is then a useful characteristic of the field define the potential difference V AB =V A -V B as the work done in moving a charge Q from point B to point A in the field E divided by Q the order of integration will take care of the signs: V A >V B if we need to do some work to bring a positive charge from B to A V AB a SCALAR quantity!

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