Two electric dipoles of moment P and 64 P are placed in opposite direction on a line at a distance of 25 cm. The electric field will be zero at point between the dipoles whose distance from the dipole of moment P is

5 cm

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 cm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 5 cm
Suppose neutral point N lies at a distance x from dipole of moment $ρ$ or at a distance $X_{2}$ from
dipole of 64 $ρ .$

$\Rightarrow \frac{1}{4 π ϵ_{0}} \frac{2 p}{x^{3}} = \frac{1}{4 π ϵ_{0}} . \frac{2 (64 P)}{(25 - x)^{3}}$
$\Rightarrow \frac{1}{x^{3}} = \frac{64}{(25 - x)^{3}} \Rightarrow x = 5 c m .$

Suggest Corrections

Similar questions

Q. Two short electric dipoles of moment

P

and

27 P

are placed as shown below. If at the point

O

the electric field is zero between the dipoles, then the distance of the point

O

from the centre of first dipole is

Q. Two short electric dipoles are placed as shown (

r

is distance between their centres). The energy of electric interaction between these dipoles will be:
(

C

is centre of dipole of moment

P_{2}

)

Q. Electric potential at equatorial point of a small dipole with dipole moment P(At, r distance from the dipole) is

Two electric dipoles each of dipole moment $p = 6.2 \times 10^{^{- 30}} C - m$ are placed with their axis along the same line and their centres at a distance d $= 10^{- 8} c m$ . The force of attraction between dipoles is:

Q. The electric field at a point on equatorial line of a dipole and direction of the dipole moment in direction.

Join BYJU'S Learning Program

Two electric dipoles of moment P and 64 P are placed in opposite direction on a line at a distance of 25 cm. The electric field will be zero at point between the dipoles whose distance from the dipole of moment P is

The correct option is A 5 cmSuppose neutral point N lies at a distance x from dipole of moment ρ or at a distance X2 from dipole of 64 ρ. ⇒14πϵ02px3=14πϵ0.2(64P)(25−x)3 ⇒ 1x3=64(25−x)3⇒ x=5cm.