Question

A dipole with a positive charge of 2.0 uC and a negative charge of -2 uC is centered at the origin and oriented along the x axis with the positive charge located to the right of the origin. The charge separation is 0.0010 m. Find the electric field due to this dipole at the point x = 4.0 m, y = 0.0 m.
(A) 0.56 i N/C
(B) -.56 i N/C
(C) 0.28 i N/C
(D) -0.28 i N/C

Answer 1

Answer:

The reulting electric field at x = 4.0 and y = 0.0 from the dipole is 0.5612 N/C

Solution:

As per the question:

Charges of the dipole, q = $\pm 2\mu C$

Separation distance between the charges, d = 0.0010 m

Separation distance between the center and the charge, d' = $\frac{d}{2} = 5\times 10^{- 4} m$

x = 4.0 m

y = 0.0 m

Now,

The electric field due to the positive charge on the right of the origin:

E = $k\frac{q}{(d' + x)^{2}}$

where

k = Coulomb's constant = $9\times 10^{9} Nm^{2}C^{- 2}$

Now,

E = $(9\times 10^{9})\frac{2\times 10^{- 6}}{(5\times 10^{- 4} + 4)^{2}} = 1124.72\ N/C$

Similarly, electric field due to the negative charge:

E' = $k\frac{q}{(x - d')^{2}}$

E' = $(9\times 10^{9})\frac{2\times 10^{- 6}}{(4 - 5\times 10^{4})^{2}} = - 1125.28\ N/C$

Thus

$E_{total} = E' - E = 0.5612 N/C$