Three identical dipoles are arranged as shown below. what will be the net electric field at P $(k = \frac{1}{4 π ε_{0}})$

\frac{k p}{x^{3}}

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

\frac{2 k p}{x^{3}}

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

Z e r o

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

\frac{\sqrt{2} k p}{x^{3}}

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

Open in App

Solution

The correct option is C $Z e r o$
In this diagram, the equilibrium the picture says that equilibrium condition. Hence at $P$ , Electric field $= 0$ .

Suggest Corrections

Similar questions

Q. Three identical dipoles are arranged as shown below. What will be the net electric field at P?

(K = \frac{1}{4 π ϵ_{0}})

Q. Two identical electric dipoles are arranged on

x

-axis as shown in figure. Electric field at the origin will be ( Assume

\frac{1}{4 π ϵ_{0}} = K

)

Q. Two identical electric dipoles are arranged on

x

-axis as shown in figure. Electric field at the origin will be ( Assume

\frac{1}{4 π ϵ_{0}} = K

)

AT temperature T, a compound $A B_{2} (g)$ dissociates according to the reaction,
$2 A B_{2 (g)} ⇌ 2 A B_{(g)} + B_{2 (g)}$ with a degree of dissociation, x (where x << 1). If KP and P are equilibrium constant and equilibrium pressure respectively, the value of x is

Q. Two short electric dipoles are placed as shown in figure.

The energy of electric interaction between these dipoles will be :

[Take k = \frac{1}{4 π ε_{0}}]

Join BYJU'S Learning Program

Three identical dipoles are arranged as shown below. what will be the net electric field at P (k=14πε0)

The correct option is C ZeroIn this diagram, the equilibrium the picture says that equilibrium condition. Hence at P, Electric field =0.

Three identical dipoles are arranged as shown below. what will be the net electric field at P $(k = \frac{1}{4 π ε_{0}})$

The correct option is C $Z e r o$
In this diagram, the equilibrium the picture says that equilibrium condition. Hence at $P$ , Electric field $= 0$ .