Question

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

• A. $\frac{k \cdot p}{x^{3}}$
• B. $\frac{2 k p}{x^{3}}$
• C. Zero
• D. $\frac{\sqrt{2} k p}{x^{3}}$

Answer 1

Answer: (C)

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 ,
$E_{1}=\frac{k \cdot p}{x^{3}}$ (towards left)
Due to dipole 2 , $E_{2}=\frac{k \cdot p}{x^{2}}$ (towards left)
Due to dipole $3, E_{3}=\frac{k(2 p)}{x^{3}}$ (towards right)
So, net field at $P$ will be zero.