5Curl-freefieldsandelectrostaticpotential • Mathematically, we can generate a curl-free vector field E(x, y, z) as ∂V ∂V ∂V E = −( , , ), ∂x ∂y ∂z by taking the gradient of any scalar function V (r)=V (x, y, z).The gradient of V (x, y, z) is defined to be the vector ∂V ∂V ∂V ∇V ≡ ( , , ), ∂x ∂y ∂z pointing in the direction of increasing V ;inabbreviatednotation,curl- free fields E can be indicated as E = −∇V. – Verification: Curl of vector ∇V is xˆ yˆ zˆ ! ∂ ∂ ∂ ! ∇×(∇V )=! ∂x ∂y ∂z ! =ˆx0 − yˆ0 − zˆ0=0. ! ∂V ∂V ∂V ! ! ! ! ∂x ∂y ∂z ! ! ! – If E = −∇V represents! an electrostatic! field,thenV is called the electrostatic potential. ◦ Simple dimensional analysis indicates that units of electro- static potential must be volts (V). 1 – The prescription E = −∇V ,includingtheminussign(optional, but taken by convention in electrostatics), ensures that electro- static field E points from regions of “high potential” to “low po- tential” as illustrated in the next example. Electrostatic fields E point from regions of “high V ” to “low V ” Example 1: Given an electrostatic potential V (x, y, z)=x2 − 6y V 4 in a certain region of space, determine the corresponding electrostatic field E = −∇V in the same region. 2 Solution: The electrostatic field is ∂ ∂ ∂ 0 E = −∇(x2 − 6y)=−( , , )(x2 − 6y)=(−2x, 6, 0) = −xˆ 2x +ˆy6 V/m. ∂x ∂y ∂z !2 Note that this field is directed from regions of high potentialtolowpotential.Alsonote that electric field vectors are perpendicular everywhere to “equipotential” contours. !4 !4 !2 0 2 4 Given an electrostatic potential V (x, y, z),findingthecorrespondingelec- Light colors indicate “high V ” trostatic field E(x, y, z) is a straightforward procedure (taking the negative dark colors “low V ” gradient) as already illustrated in Example 1. The reverse operation of finding V (x, y, z) from a given E(x, y, z) can be accomplished by performing a vector line integral o E · dl "p 2 in 3D space, since, as shown below, such integrals are “path independent” for p =(x ,y ,z ) curl-free fields E = −∇V . z p p p Ej • The vector line integral y ∆lj o C E · dl C! "p o =(xo,yo,zo) over an integration path C extending from a point p =(xp,yp,zp) in 3D space to some other point o =(xo,yo,zo) is defined to be x – the limiting value of the sum of dot products Ej·∆lj computed over all sub-elements of path C having incremental lengths |∆lj| and unit vectors ∆lj/|∆lj| directed from p towards o —thelimiting value is obtained as all |∆lj| approach zero (i.e., with increasingly finer subdivision of C into |∆lj| elements). • Computation of the integral (see example below) involves theuseof infinitesimal displacement vectors dl =ˆxdx +ˆydy +ˆzdz =(dx, dy, dz) and vector dot product E · dl =(Ex,Ey,Ez) · (dx, dy, dz)=Exdx + Eydy + Ezdz. The integral o o E · dl = (Exdx + Eydy + Ezdz) "p "p 3 will in general be path dependent except for when E is curl-free. Curl-free: path-independent line integrals Cu 1.0 Example 2: The field E =ˆxy ± yxˆ is curl-free with the + sign, but not with − as verified below by computing ∇×E.CalculatethelineintegralofE (for both 0.8 signs, ±)fromapointo =(0, 0, 0) to point p =(1, 1, 0) for two different paths C going through points u =(0, 1, 0) and l =(1, 0, 0),respectively(seemargin). 0.6 z Cl Solution: First we note that 0.4 xˆ yˆ zˆ 0.2 ∇×(ˆxy ± yxˆ )=! ∂ ∂ ∂ ! =ˆz(±1 − 1) ! ∂x ∂y ∂z ! ! y ±x 0 ! 0.0 ! ! 0.0 0.2 0.4 0.6 0.8 1.0 ! ! x E ! ! which confirms that =ˆxy ± yxˆ is curl-free with with + sign, but not with −. “Curly”: path-dependent line In either case, the integral to be performed is integrals p p p Cu 1.0 E · dl = (Exdx + Eydy + Ezdz)= (ydx± xdy). "o "o "o 0.8 For the first path Cu going through u =(0, 1, 0),wehave p 1 1 0.6 (ydx± xdy)= (±x) dy|x=0 + ydx|y=1 =0+1=1. "o "y=0 "x=0 z Cl 0.4 For the second path Cl going through l =(1, 0, 0),wehave p 1 1 0.2 (ydx± xdy)= ydx|y=0 ± xdy|x=1 =0± 1=±1. "o "x=0 "y=0 0.0 p E l path independent 0.0 0.2 0.4 0.6 0.8 1.0 Clearly, the result shows that the line integral o · d is for x E =ˆxy +ˆyx which is curl-free, and path dependent# for E =ˆxy − yxˆ in which case ∇×E =0& . 4 • The mathematical reason why curl-free fields have path-independent line integrals is because in those occasions the integrals can be written in terms of exact differentials: – for curl-free E =ˆxy +ˆyx we have E · dl as an exact differential p E l ydx + xdy = d(xy) of the function xy,inwhichcase o · d = p xy|o =(1· 1 − 0 · 0) = 1 over all paths. # – for E =ˆxy − yxˆ with ∇×E = −2ˆz &=0,ontheotherhand, E · dl = ydx − xdy does not form an exact differential −dV ,and p thus there is no path-independent integral −V |o,noranunderlying potential function V . z o E · l Vp = !p d E l E ∂V ∂V ∂V E r ·d is guaranteed to be an exact differential if = −∇V =(− ∂x , − ∂y , − ∂z ), ( ) y since in that case the differential of V (x, y, z),namely dl ∂V ∂V ∂V dV ≡ dx+ dy+ dz, is precisely −Exdx−Eydy−Ezdz = −E·dl. ∂x ∂y ∂z Vo =0 – In that case x o o p E · dl = − dV = dV = Vp − Vo "p "p "o is independent of integration path; thus, if we we call o the “ground”, and set Vo =0,then o Vp = E · dl "p denotes the potential drop from (any) point p to ground o. 5 • The physical reason why this integral formula for potential Vp works z o E · l Vp = !p d with any integration path is the principle of energy conservation: E r ( ) y dl o E l E – integral p · d represents the work done by field per unit charge #moved from p to o,soifthelineintegralwerepath- Vo =0 dependent there would be ways of creating net energy by making achargeq follow special paths within the electrostatic field E, x As long as E is curl-free, line integral is path-independent and in violation of the general principle of energy conservation(that produces the voltage drop from permits energy conversion but not creation or destruction). point p to "ground" o. z − p E · l Example 3: Given that Vo = V (0, 0, 0) = 0 and Vp = !o d V y E =2xxˆ +3zyˆ +3(y +1)ˆz , m Z determine the electrostatic potential Vp = V (X, Y, Z) at point p =(X, Y, Z) in volts. o Y Solution: Assuming that the field is curl-free (it is), so that any integration path can X x be used, we find that o p p Vp = E · dl = − E · dl = − (2xdx+3zdy+3(y +1)dz) "p "o "o X Y Z = − 2xdx|y,z=0 − 3zdy|x=X,z=0 − 3(y +1)dz|x=X,y=Y "0 "0 "0 2 = −X − 0 − 3(Y +1)Z. This implies 2 V (x, y, x)=−x − 3(y +1)z V. 6 Note that 2 2 −∇(−x − 3(y +1)z)=∇(x +3(y +1)z) =ˆx2x +ˆy3z +ˆz3(y +1) yields the original field E,whichisanindicationthatE is indeed curl-free. Example 5: According to Coulomb’s law electrostatic field of a proton with charge Q = e (where −e is electronic charge) located at the origin is given as z E e = 2 r,ˆ 4π#or where 2 2 2 (x, y, z) z = r r = x + y + z and rˆ = . $ r Determine the electrostatic potential field V established by charge Q = e with the provision that V → 0 as r →∞(i.e., ground at infinity). Solution: Field E and its potential V will exhibit spherical symmetry in this problem. Therefore, with no loss of generality, we can calculate the line integral from a e y point p at a distance r from the origin to a point o at ∞ (the specified ground) x along, say, the z-axis. Approaching the problem that way, the potential drop from r to ∞ is ∞ e V (r)= 2 zˆ · zdzˆ "z=r 4π#oz e ∞ e = − |r = . 4π#oz 4π#or 7 • To convert electrostatic potential Vp (in volts) at any point p to poten- tial energy of a charge q brought to the same point, it is sufficient to multiply Vp with q (or just the sign of q,dependingonwhichenergy units we want to use — see the next example). Example 6: In view of Example 5, what are the potential energies of a proton e and an electron −e placed at distance r = a away from the proton at the origin, where distance !2 4π#o −10 z a ≡ 2 =0.529 × 10 m e me stands for Bohr radius —itisthemeandistanceofthegroundstateelectronin 19 ahydrogenatomfromthecenteroftheatom.Recallthate =1.602 × 10− C ±e z = a −9 and #o ≈ 10 /36π F/m. Solution: Let’s first evaluate the potential V (r) at r = a: e (1.6 × 10−19)36π × 109 9 × 1.6 V (a)= ≈ −10 = =27.2 V. 4π#oa 4π × 0.53 × 10 0.53 e y For the proton, potential energy in Joules is calculated by multiplying V (a)=27.2 x Vwithq = e =1.602 × 10−19 C.