AP EAPCET 23-Aug-2021 Shift 2 Paper

Consider two dipoles are placed at two points A and B in opposite direction.Let P be the null point, where net electric field intensity is found to be zero.
x be distance of point from electric dipole at A. By using expression of electric field intensity at axial position due to short electric dipole,
E=

2kp

r3

E1 and E2 be the magnitudes of electric field due to electric dipole p and 27p. which are equal and opposite to produce net electric field zero.
Therefore, E1=E2

2kp

x3

=

2k(27p)

(24−x)3

⇒

1

x

=

3

24−x

24−x=3x⇒24=4x
⇒x=6‌cm from point A