equipotential surface

the e field ina region is given by E=4axy√zi+2ax2√zj+ax2y/√zkwhere a is apositive constant eqn of equipotetial surface
(a)z=c/(x3y2)

(b)z=c/(xy2)
(c)z=c/x4y2)
(d)none
here c is a constant

3 Answers

33
Abhishek Priyam ·

nice one...

U=-E.dr
=-∫(4axy√z i + 2ax2z j + ax2y/√z k)(dx i+dyj+dzk)

=-∫(4axy√z dx + 2ax2z xy + ax2y/√z dz)
U=-0,0,0∫x,y,zd(2ax2y√z)
ΔU=-2ax2y√z=K(constant)

z=K/(2ax2y)
z==c/x4y2 ...c=k2/4a2

option (c)

1
vector ·

correct:D

33
Abhishek Priyam ·

whats there for :D

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