Two charges of equal magnitude 2 μC but opposite in sign are separated by a distance of 16 cm. Calculate the electric field at an equatorial point which is at a distance of 18 cm away from the center of the dipole

CONCEPT: 

• The electric field for the points on the equatorial plane is given by

​\(⇒ E = -\frac{P}{4\pi\epsilon_{0}(r^{2}+a^{2})^{\frac{3}{2}}}\)

• if the value of r >>>>a the electric field is given by

\(⇒ E = -\frac{P}{4\pi\epsilon_{0}r^{3}}\)

Where P = Diploe momentum, r = Distance, and a = Half of distance between both the charge.

CALCULATION:

Given - q = 2μC = 2 × 10^-6 C, 2a = 16cm = 16 × 10^-2 m , a = 8 × 10-2 m, r = 18 cm =  18 × 10-2 m, \(\frac{1}{4\pi\epsilon_{0}} = 9× 10^{9}\)

• The diploe momentum can be calculated as

⇒ P = 2a × q = 2 × 10^-6 × 16 × 10^-2 = 32× 10^-8 C-m

• The electric field for the points on the equatorial plane is given by

​\( ⇒ E = -\frac{P}{4\pi\epsilon_{0}r^{3}}\)

Substitute the given values in the above equation

\(⇒ E = -\frac{9× 10^{9}× 32× 10^{-8}}{(18\times 10^{-2}) ^{3}}=\frac{4\times 10^{7}}{81}=0.049\times10^7=49\times 10^4 NC^{-1}\)

• Hence option 2 is the answer