Problem 2: For the point fp=3, p=60°,z=2) in cylindrical coordinates and the potential field V= 10(p+1)z'cos Vin free space, find at P:a) (b) E; (c) D; (d) e (a) (b) E; (c) direction of Ep; (d) D; (e) pr at P. Solution: (a) At point P: Vp = =___? Av + papan + a = ?Axt ? Ayt ?A, 1 av av (b) 2 = - = ap (c) EP At point P: Ep = ___?Ax +_?Ay +_ _?A, EP =__?A, +_?A, +-_?A, Epl (d) D = EE = _?Ax + ?A, + ? A, At point P: Dp = pc = _?Ax + _?A, +_?A. (e) Py = div D = div , E == (8.854 x 10-17 ?) =_? At P: Py =