Adipole 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

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 =

Separation distance between the charges, d = 0.0010 m

Separation distance between the center and the charge, d' =

x = 4.0 m

y = 0.0 m

Now,

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

E =

where

k = Coulomb's constant =

Now,

E =

Similarly, electric field due to the negative charge:

E' =

E' =

Thus

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

that the claims meet official rules.

don't misinterpret the claims