Answer to Question #213526 in Electricity and Magnetism for Muthiah Afifah

Question #213526

A potential field is given by V = 3x^2 y - yz. Which of the following is not true?

(a) At point (1, 0, -1) V and E vanish.

(b) x^2 y = 1 is an equipotential line on the xy-plane.

(c) The equipotential surface V= -8 passes through point P+2, -1, 4

. (d) The electric field at P is 12ax - 8ay - az V/m.

(e) A unit normal to the equipotential surface V =-8 at P is -0.83ax + 0.55ay + 0.07az


1
Expert's answer
2021-07-12T03:10:01-0400

V=3x2yyzV=3x^2y-yz

V(1,0,1)=0V_{(1,0,1)}=0

E=VE=-\nabla V


E1,0,1=(6(0)ax+(3×1+1)ay0az)E_{1,0,1}=-(6(0)a_x+(3\times1+1)a_y-0a_z)

E=4ayE=-4a_y

Option (a) wrong(incorrect)

P=x2y1P=x^2y-1

Equpotential line equation


.P=.(x2y1)=2xyax+x2ay(2)\nabla.P=\nabla.(x^2y-1)=2xya_x+x^2a_y\rightarrow(2)

Equation (2) is lies xy-plane


Option (b) is correct option


Electric field

V=3x2yyzV=3x^2y-yz


V(2,1,4)=3×4(1)(1)×4=8VV_{(2,-1,4)}=3\times4(-1)-(-1)\times4=-8V

Option (c) is correct


E=V=(6xyax+(3x2z)ayyaz)E=-\nabla V=-(6xya_x+(3x^2-z)a_y-ya_z)

E=12ax8ayazE=12a_x-8a_y-a_z

Option (d) is correct

E=12ax8ayazE=12ax-8ay-az

V^=VV\hat{V}=\frac{\nabla V}{ |\nabla V|}


V^=6axyax+(3x2z)ay+az(y)14.5\hat{V}=\frac{6axy a_x+(3x^2-z)a_y+a_z(-y)}{14.5}

V^(2,1,4)=1214.4ax+814.4ay+114.4az\hat{V}_{(2,-1,4)}=\frac{-12}{14.4}a_x+\frac{8}{14.4}a_y+\frac{1}{14.4}a_z

V^=0.83ax+0.55ay+0.07az\hat{V}=-0.83a_x+0.55a_y+0.07a_z

Option (e) is correct


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment