The equation of an equipotential line in an electric field is y=2x, then the electric field strength vector at (1,2) may be :
4^i+3^j
4^i+8^j
8^i+4^j
−8^i+4^j
A
4^i+8^j
B
8^i+4^j
C
4^i+3^j
D
−8^i+4^j
Open in App
Solution
Verified by Toppr
Now equation of equipotential surface is y=2x Now electric field along the euipotential surface should be zero therefore angle made by equipotential surface with x-axis is tan−1(2) Now since net electric field should be perpendicular to the equipotential surface therefore for any electric field which makes an angle tan−1(−1/2) with x-axis can be the electric field at point (1,2) which is true only for option (D)
because for two perpendicular line, product of their slope should be equal to -1 i.e., m1×m2=−1
Was this answer helpful?
4
Similar Questions
Q1
The equation of an equipotential line in an electric field is y=2x, then the electric field strength vector at (1,2) may be :
View Solution
Q2
The equation of an equipotential line in an electric field is y=2x, then the electric field strenght vector at (1,2) may be:
View Solution
Q3
The equation of an equipotential region (line) is y=3x4, then electric field vector at the origin may be
View Solution
Q4
Unit vector parallel to the resultant of vectors →A=4^i−3^j and →B=8^i+8^j
View Solution
Q5
Moment of the force, →F=(4^i+5^j−6^k)N at P(2,0,−3)m about the point Q(2,−2,−2)m is given by