Question : Two point charges of 5µC but in opposite sign are placed 4 cm apart. Calculate the electric field intensity at a point 4 cm from the midpoint on the axial line of the dipole.
Doubt by Prateek
Solution :
Here
q = ±5µC
q = 5×10-6 C
2a = 4cm
2a = 4×10-2 m
a = 2×10-2 m
a = 0.02
r = 4 cm
r = 4×10-2 m
r = 0.04 m
Dipole Moment (p)
= q×2a
= 5×10-6×4×10-2
= 20×10-8 Cm
If you properly read the question then you will realise that we have to calculate the electric field at an axial point of the dipole.
Here r=0.04 m and a=0.02 m
So we will not use the formula for short dipole.
We know, electric field due to a dipole at any point on the axial axis of the dipole is given by
E = 2kpr(r²-a²)²
E = [2×(9×109)(20×10-8)(0.04)]/[(0.04)²-(0.02)²]²
E = [2×9×20×0.04×109-8]/[0.0016-0.0004]²
E = [144/(0.0012)²]
E = (144/144)×108
E = 108 N/C
