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- The electric field is conservative, so
**F**= - ∇ U = - q_{t}∇ V.**electric field**(**Coulomb's law**):**E**(x,y,z) ≡ - ∇ V(x,y,z)= (1 / (4 π ε)) ∫

**r**dq / r^{3}**r**in the integrand is the vector pointing*from*the charge dq to the field point. The integrand is most often written as q times the unit vector pointing in the r direction, divided by r^{2}, but this expression is more useful unless the problem is 1-dimensional. Either way, the field (and therefore the force) decreases as 1/r^{2}.Note that the electric field has units both of N / C and V / m. Think of it both as a force per unit charge and a potential per unit distance.

The direction of the electric field at a point is defined as the direction of the acceleration experienced by a positive test charge:

Since the electric field is minus the gradient of the potential field,

ΔV = - ∫

where the integral is along a path between two field points.**E ⋅ ds**,Consider again the line of charge extending from -b to b. We could compute the electric field using

**E**(x,y) = (λ / (4 π ε)) ∫_{-b}^{b}(x - s, y) ds / (√ ((x - s)^{2}+ y^{2}))^{3}**E**(x,y) = - ∇ V(x,y)= (λ / (4 π ε)) (1 / √((b - x)

^{2}+ y^{2}) - 1 / √((b + x)^{2}+ y^{2}),((b - x) / y) / √((b - x)

^{2}+ y^{2}) + ((b + x) / y) / √((b + x)^{2}+ y^{2}))**E**(y) = λ b / (2 π ε y √ (b^{2}+ y^{2}))**j****E**(x,y) = λ / (2 π ε y)**j** - Always check to see if one or more components of the electric field are zero because of symmetry of the source charge configuration.
- When the charges are in equilibrium, the electric field is zero inside a conductor, and normal to its surface.

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©2010, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.

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