f,E.ds_60The left hand side of the above equationisf,E.ds=f,E-e,ds-JEdS=4rr'EAnd we haveqqEEor4元r24元8We also can use the formula for electric potential or electricfield intensity to calculate the electric field intensity produced bythe point charge.U7
The left hand side of the above equation is = = = S S S E S r E 2 E dS E en dS d 4 And we have 2 4π 0 r q E = r r q E e 2 π 0 4 or = = S q 0 d E S We also can use the formula for electric potential or electric field intensity to calculate the electric field intensity produced by the point charge
If the point charge is placed at the origin, then r - r'I- r. We findthe electric potentialproducedby the point charge asqp(r) =4元80rTheelectricfieldintensityEas9E=-Vβ=4元804元6,rIf the eguation for electric field intensityis used directly, we findtheelectricfieldintensityEaso(rqE:dv4元804元6.U7
If the point charge is placed at the origin, then . We find the electric potential produced by the point charge as |r − r |= r r q 4π 0 ( ) r = r r q r q E e 2 0 4π 0 1 4π = = − = − The electric field intensity E as r V r r q V r e r e E 2 0 2 0 4π d 4π ( ) = = If the equation for electric field intensity is used directly, we find the electric field intensity E as
Example2Calculatetheelectricfield intensityproducedbyan electric dipole.Solution: For an electric dipole consistsoftwo point electric charges, we can use theprinciple of superposition to calculate theelectricfieldintensity.Then the electric2potentialproduced by an electric dipole asqaP:4元20+4元04元.rIf the distancefrom the observeris much greater than the separationl, we can considerthate, and e, are parallel with e., andcoser+=cosorr. =r-1r. -r =lcoseU2V
Example 2 Calculate the electric field intensity produced by an electric dipole. Solution: For an electric dipole consists of two point electric charges, we can use the principle of superposition to calculate the electric field intensity. Then the electric potential produced by an electric dipole as − = − = + − − + + − r r q r r r q r q 4π 0 4π 0 4π 0 If the distance from the observer is much greater than the separation l, we can consider that and are parallelwith , and + r e − r e r e r− − r+ = l cos 2 cos 2 cos 2 r l r l r r r + + − = − x -q +q z y l r r- r+ O
q9l cos0 =(l.er)WehaveP4元604元80where the direction of the vector l is defined as the directionpointing to the positive charge from the negative charge. Theproduct ql is often called the electric moment of the dipole, and isdenoted by p, so thatp= qlThen the electric potentialofthe dipole can be written aspcosap.e.4元8024元20rU7
where the direction of the vector l is defined as the direction pointing to the positive charge from the negative charge. The product ql is often called the electric moment of the dipole, and is denoted by p,so that p = ql ( ) 4π cos 4π 2 0 2 0 r r q l r q = = l e We have 2 0 2 0 4π cos 4π r p r r = = p e Then the electric potential of the dipole can be written as
By using the relation E --Vo, we can find the electric fieldintensity of the electric dipole as1#pcosoa100appsin eE=ee七4元50r2元80arr a0rsin 0 apThe electric potentialof the dipole is inversely proportionalto thesquare of the distance, and the magnitude of the electric field intensityis inversely proportional to the third power of the distance.In addition, the electric potentialand the electric field intensityare both dependent of the azimuthal angle .These properties are very different from that of a point charge.UV
+ + = − sin 1 1 r r r E r e e e 3 0 3 0 4π sin 2π cos r p r p r = e + e By using the relation , we can find the electric field intensity of the electric dipole as E = − * The electric potential of the dipole is inverselyproportional to the square of the distance, and the magnitude of the electric field intensity is inversely proportional to the third power of the distance. These properties are very differentfrom that of a point charge. * In addition, the electric potential and the electric field intensity are both dependent of the azimuthal angle