Two short dipoles (A, B), A having charges ±2μC and length 1 cm and B having charges ±4μC and length 1 cm are placed with their centres 80 cm apart. The electric field at a point P equidistant from the centres of both dipoles is ___ N/C.
Q. Two short dipoles $(A,B)$, $A$ having charges $\pm2\mu C$ and length $1\,cm$ and $B$ having charges $\pm4\mu C$ and length $1\,cm$ are placed with their centres $80\,cm$ apart. The electric field at a point $P$ equidistant from the centres of both dipoles is ____ $\text{N/C}$.
A. $\dfrac{9}{16}\sqrt{2}\times10^{5}$
B. $\dfrac{9}{16}\sqrt{2}\times10^{4}$
Correct Answer: $\dfrac{9}{16}\sqrt{2}\times10^{4}$

Explanation

The centres of the two dipoles are $80\,cm$ apart and point $P$ lies exactly at the midpoint. Hence distance of $P$ from each dipole centre is

$$ r = 40\,cm = 0.4\,m $$

Dipole moment of a short dipole is

$$ p = ql $$

For dipole $A$:

$$ p_A = (2\times10^{-6})(1\times10^{-2}) = 2\times10^{-8}\,C\cdot m $$

For dipole $B$:

$$ p_B = (4\times10^{-6})(1\times10^{-2}) = 4\times10^{-8}\,C\cdot m $$

Electric field due to a short dipole at a general point varies as

$$ E \propto \frac{p}{r^3} $$

Thus,

$$ E_A : E_B = p_A : p_B = 1 : 2 $$

Using $k = 9\times10^9$,

$$ E_A = \frac{9\times10^9 \times 2\times10^{-8}}{(0.4)^3} = 5.625\times10^3\,N/C $$

$$ E_B = \frac{9\times10^9 \times 4\times10^{-8}}{(0.4)^3} = 11.25\times10^3\,N/C $$

The two dipoles are perpendicular, hence electric fields at $P$ are also perpendicular.

$$ E = \sqrt{E_A^2 + E_B^2} $$

$$ E = \frac{9}{16}\sqrt{2}\times10^4\,N/C $$

Related JEE Main Physics Questions

Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.

Scroll to Top