Question

where c> 0 ro The electric field in the xy-plane due to an infinite line of charge along the z-axis is a gradient field with a potential function V(x,y)=c In Vx2 + y2 is a constant and ro is a reference distance at which the potential is assumed to be 0. Use this information to answer parts a through c. wherer= x2 + y2. Rewrite E in terms b. Show that the electric field at a point in the xy-plane is directed outward from the origin and has magnitude |E| = of r. Choose the correct answer below. OA. E. Irl? OBEC rolul O C. E cro 1812 с OD. EN lo Rewrite E again. Choose the correct answer below. OA. E с г ro OB. E cr ro co OC EE с г OD. E= Therefore, by definition, the electric field has magnitude |EI = c. Show that the vector field is orthogonal to the equipotential curves at all points in the domain of V. Plot the equipotential curves of V. What are the resulting shapes? o Parabolas O Ellipses 0 Circles O Hyperbolas What is a vectort that is always tangent to this shape? O A. t= (-x-Y) OB. t= (-y.-) OC. t= (-x,y) OD. t= (-y.x) Find the dot product of E and t. What is the result? 0 (Simplify your answer.) Therefore, by definition, the vector field is orthogonal to the equipotential curves at all points in the domain of V.

Answer 1

Given data; Vex y) = c Informations where cro reference are constants, to is some ro and Co distance so 5x²142 EE х 3 (b) The electric field is defined as: Ē = -7 voorry) E = -(20 m to boy ) (eln Take x-combo c/x²+42 8o(-1/2x (Jaya) 3 с (a442) where r=1r1= 5x²442 a En = cni similarly Ey= 7? so CI 12 1812 7 Because ²= runit vector 2 E- & away from the centre, *Hence the electric field points outward away from the origin. * option @ E = TETE ř is the correct answer z E cy 5 - Ē = {(ai+yî) 22 82

from part (6) Ê cr 1512 ใน) where f = 1 са Hence E = u visi g is the corect answer oftion # (D) and the magnitude of electric field is TEI= Fr because 7 Tri ř is just the unit rector (C) At the equipolential curves V(xy=k K io seme constant, so: In =k en snya) = = klo Cr no √x Yeyz 7. su ē $77. Take mko ſx442 = T some constant because this term is a que contant

on both sides 1 Hence √x²tyzat squasing ty? This is the equation of a circle of radius to take this curve as $= x+y 27? Normal to this surface is given by ñ = 54 = 22 i + 2y 9 nity 1581 (2294 (24)2 5x²+42 17 त We see that the normal to this equipotential in the direction of electric - n = curve is field E = 6 181 18 + Hence, Ē je orthogonal to the equibotential proved curves. curues is And from egn the shape of these circles. Hence option 3rd CIRCLES is connect.

れこ We have to choose a vector E that is always tangent to this shape- meaning x+y=7? We have seen that the normal to this shape was xi+ y √x²+y2 Hence, te vector Ę must be perpendiculas to ñ in order to be tangent to this curve. Ao sin (A) => --> La Care because 7 = -x²-y2 70 Jazry 2 (B) F=<4-20 is also incorrect because - {xy -224) E.T- -2xy √x?g? (x2+y2 70 (C) #= (-2,y) is also incorrect {={-4, n> is correct È t = -yxt ay 20 ED be cause Jx²+y2 so Ē. Tro Ciin and option (P2 is the comect answer. Dot product is zeso from (il,