Three identical, small dipoles are arranged as shown below. What will be the net electric field at P ?(k=(1/4 π ε0))

Question

Three identical, small dipoles are arranged as shown below. What will be the net electric field at P ?(k=(1/4 π ε0)) (A) (k ⋅ p/x3) (B) (2 k p/x3) (C) Zero (D) (√2 k p/x3)

Tardigrade Admin · Accepted Answer

Point P lies at equatorial positions of dipole 1 and 2 and axial position of dipole 3 . Hence, field at P due to dipole 1 , E1=(k ⋅ p/x3) (towards left) Due to dipole 2 , E2=(k ⋅ p/x2) (towards left) Due to dipole 3, E3=(k(2 p)/x3) (towards right) So, net field at P will be zero. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0f17803e6ecbc0d1dfde77384a0fc775-.png" />

Q. Three identical, small dipoles are arranged as shown below. What will be the net electric field at $P ? (k = \frac{1}{4 π ε _{0}})$

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

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

Zero

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

Q. Three identical, small dipoles are arranged as shown below. What will be the net electric field at P?(k=4πε0​1​)

x3k⋅p​

x32kp​

Zero

x32​kp​

Point P lies at equatorial positions of dipole 1 and 2 and axial position of dipole 3. Hence, field at P due to dipole 1 , E1​=x3k⋅p​ (towards left) Due to dipole 2 , E2​=x2k⋅p​ (towards left) Due to dipole 3,E3​=x3k(2p)​ (towards right) So, net field at P will be zero.

Q. Three identical, small dipoles are arranged as shown below. What will be the net electric field at $P ? (k = \frac{1}{4 π ε _{0}})$

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

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

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