Electricfield linesandequipotentialsurfacesofan electricdipoleElectric field linesEquipotential surfacesUV
Electric field lines Electric field lines and equipotential surfaces of an electric dipole Equipotential surfaces
Example 3Assume that an infinitelylong charged cylinder of radiusa is placedin free space. The density ofthe charge is p. Calculate theelectric field intensities inside and outsidethe cylinderSolution: Choose a cylindrical coordinatesystem. Since the cylinder is infinitely long,for any z the environmentremain the sameHence the field is independent of thecoordinatevariablez.Sinceitis symmetricalabout any planeof z = constant,the electric field intensitymust be perpendicular to the z-axis, andcoplanarwith the radialcoordinater.In addition, the cylinderis rotationally symmetrical.Thus the fieldmust be independent of the angle Φ
Example 3 Assume that an infinitely long charged cylinder of radius a is placed in free space. The density of the charge is . Calculate the electric field intensities inside and outside the cylinder. x z y a L S1 Solution: Choose a cylindrical coordinate system. Since the cylinder is infinitely long, for any z the environment remain the same. Hence the field is independent of the coordinate variable z. In addition, the cylinder is rotationally symmetrical. Thus the field must be independent of the angle . Since it is symmetrical about any plane of z = constant, the electric field intensity must be perpendicular to the z-axis, and coplanarwith the radial coordinate r
Construct a cylindrical Gaussian surface of radius r and length LApplying Gauss'law, we havef.E·ds-q80Since the direction of the electric field intensity coincides withthe outward normal of lateral surface S, of the cylinder everywhereand is perpendicular to the normal to the upper or the lower endfaces, the left side of the surface integral becomesfE.dS-J,EdS =EJdS=2rLEFor r < a , the charge q = πr? pL , and the electric field intensityisE=pr280U7
Since the direction of the electric field intensity coincides with the outward normal of lateral surface S1 of the cylinder everywhere and is perpendicular to the normal to the upper or the lower end faces,the leftside of the surface integral becomes E S E S rLE S S S d d d 2π 1 1 = = = E S For r < a , the charge , and the electric field intensity is q r L 2 = π r r E e 2 0 = Construct a cylindrical Gaussian surface of radius r and length L. Applying Gauss’ law, we have = S q 0 d E S
When r > a, the charge q = πa pL, we have元a~OE2元60rwhere xa'p can be considered as the charge per unit length, sothat the electric field can be thought of as caused by a line chargewith densityp, = a'p. In view of this we can derive the electric fieldintensity caused by an infinite line charge with line density p, asOE2元80rFor this example, it is easy to calculate the electric field intensityusing Gauss'law. If we directlycalculate the electric fieldintensityfrom the distribution of the charge, it will be very complicatedU
r l r E e π 0 2 = For this example, it is easy to calculate the electric field intensity using Gauss’ law. If we directly calculate the electric field intensity from the distribution of the charge, it will be very complicated. When r > a, the charge , we have q a L 2 = π r r a E e 0 2 2π π = where can be considered as the charge per unit length, so that the electric field can be thought of as caused by a line charge with density . In view of this we can derive the electric field intensity caused by an infinite line charge with line density as 2 l = πa 2 πa l
Example 4Find the electric field intensity caused by a linecharge with length L and charge density p,Solution: Let the z-axis of a cvlindrical元0P(rZcoordinate system coincidewiththeline2charge and the mid point of the line beat the origin.Since it is rotationally symmetrical.the field is independent ofthe angle @It is impossible to apply Gauss'law to calculate the field becausethe direction ofthe electric field intensity cannot be found out inadvance. We have to evaluate directly the electric potentialand theelectricfieldintensity.U
x z y r 2 1 r O r − r dz z r z e r e , ) 2 π P(r, z Example 4 Find the electric field intensity caused by a line charge with length L and charge density . l Solution: Let the z-axis of a cylindrical coordinate system coincide with the line charge and the mid point of the line be at the origin. It is impossible to apply Gauss’ law to calculate the field because the direction of the electric field intensity cannot be found out in advance. We have to evaluate directly the electric potential and the electric field intensity. Since it is rotationally symmetrical, the field is independent of the angle