Home
Class 12
PHYSICS
An electric dipole lie at the origin (0,...

An electric dipole lie at the origin (0,0,0) with dipole moment `vecp=(-hati-3hatj+2hatk)xx10^(-29)`Cm. The electric field due to this dipole at `vecr=(+hati+3hatj+5hatk)` (note that `vecr*vecp=0`) is parallel to:

A

`(+hati +3hatj -2hatk)`

B

`(-hati+3hatj-2hatk)`

C

`(+hati-3hatj-2hatk)`

D

`(-hati-3hatj+2hatk)`

Text Solution

Verified by Experts

The correct Answer is:
A
since `vecp*vecr=0`
`vecE` must be antiparallel to `vecp`
So, `vecE= -lambda(vecp)`
where `lambda` is a arbitrary positive constant
Now, `vecA=a hati +b hatj +c hatk`
`vecA||vecE`
`(a)/(lambda)=(b)/(2lambda)= (c )/(-2lambda)=k`
so `vecA= lambdak (hati +3hatj -2hatk)`
Promotional Banner

Topper's Solved these Questions

  • NTA TPC JEE MAIN TEST 55

    NTA MOCK TESTS|Exercise PHYSICS (SUBJECTIVE NUMERICAL)|10 Videos

  • NTA TPC JEE MAIN TEST 54

    NTA MOCK TESTS|Exercise PHYSICS|30 Videos

  • NTA TPC JEE MAIN TEST 56

    NTA MOCK TESTS|Exercise PHYSICS |30 Videos

Similar Questions

Explore conceptually related problems

An electric dipole of moment vecP=(hati+3hatj-5hatk)xx10^(-19)Cm is kept in a electric field vecE=(2hati-3hatj+2hatk)xx10^(5)V/m . Find torque experienced by dipole.

An electric dipole with dipole moment vec p=(3hati +4hatj) C-m is kept in electric field vec E=0.4hatk N/C .What is the torque acting on it and the potential energy of the dipole?

A dipole of moment vecp = 10^(-7) (5hati + hatj - 2hatk) C is placed in an electric field vecE = 10^(7) (hati + hatj + hatk) V m^(-1) . Find the torque experienced .

A tiny electric dipole of dipole moment vecP = P_0hatj is placed at point (l,0). There exists an electric field vecE = 2ax^2hati + (2by^2 + 2cy)hatj .

The direction cosines of the perpendicular from the origin to the plane vecr.(6hati-3hatj-2hatk)+1=0 are

Find the length of perpendicular from the origin to the plane vecr.(2hati-3hatj+6hatk)+14=0 .

Show that the line vecr=(2hati-2hatj+3hatk)+lambda(hati-hatj+4hatk) is parallel to the plane vecr.(hati+5hatj+hatk)=7 .

Find the direction cosines of the perpendicular from the origin to the plane vecr.(6hati-3hatj-2hatk)+1=0 .

The line of intersection of the planes vecr . (3 hati - hatj + hatk) =1 and vecr. (hati+ 4 hatj -2 hatk)=2 is: