Exercise:
Show that the electric field can be derived as the gradient of the corresponding potential for abcliste abc a spherical charge distribution with Vx y z k_C fracQr k_C fracQsqrtx^+y^+z^ abc a linear charge distribution along z with Vx y z fraclambdapi varepsilon_ lnfracr_r fraclambdapivarepsilon_ lnfracr_sqrtx^+y^ abcliste

Solution:
abcliste abc For the x component we find E_x -fracpartial Vx y zpartial x -fracpartialpartial x left k_C fracQsqrtx^+y^+z^ right -k_C Q fracpartialpartial x left x^ + y^ + z^ right^-/ -k_C Q frac- left x^ + y^ + z^ right^-/ x k_C Q fracxleft x^+y^+z^ right^/ k_C Q fracxr^ The y and z components can be found in the same way. The electric field vector can thus be written as vec Ex y z k_C fracQr^ pmatrix x y z pmatrix With vec r pmatrix x y z pmatrix and hat r fracvec rr this is the same as Evec r k_C fracQr^ hat r which is the vector expression for the field at distance r from a spherical charge distribution Q. abc For the x component we find E_x -fracpartial Vx y zpartial x -fracpartialpartial x left fraclambdapivarepsilon_ lnfracr_sqrtx^+y^ right -fraclambdapivarepsilon_ fracsqrtx^+y^r_ r_ frac- left x^+y^ right^-/ x fraclambdapivarepsilon_ fracxx^+y^ The y component can be derived in the same way: E_y fraclambdapivarepsilon_ fracyx^+y^ For the z component we find E_z -fracpartial Vx y zpartial z -fracpartialpartial z left fraclambdapivarepsilon_ lnfracr_sqrtx^+y^ right since the potential does not dep on z. abcliste With vec r pmatrix x y pmatrix and hat r fracvec rr we can write the electric field vector at a distance r from the linear charge distribution as vec Er fraclambdapivarepsilon_ r hat r which corresponds to the expected result i.e. a field perpicular to the direction of the linear charge distribution i.e. vec r perp hat z which decreases with /r.