A−12 nc charge is located at (x, y) = (1.0 cm, 0 cm).what is the electric field at the position (x₁,y₁) = (5.0 cm, 0 cm) in component form? what is the electric field at the position (x₂,y₂) = (−5.0 cm, 0 cm) in component form? what is the electric field at the position (x₃,y₃) = (0 cm, 5.0 cm) in component form? express your answer in terms of the unit vectors i^ and j^. use the 'unit vector' button to denote unit vectors in your answer.

Explanation:

Electric field due to a charge q

= k x q / d² where k is a constant equal to 9 x 10⁹ , q is given charge and d is distance of point from the charge where field is to be measured. Direction of electric field is towards the force that the charge applies on unit positive charge at the given point.

the electric field at the position (x₁,y₁) = (5.0 cm, 0 cm)  

here , q = - 12 x 10⁻⁹ C , d = 5-1 = 4  cm

E = 9 X 10⁹ x - 12 x 10⁻⁹ /( 4 x 10⁻²)²

= - 6.75  x 10⁴ N/C

It will act towards the origin along - x axis.

E =   - 6.75 x 10⁴ i

the electric field at the position (x₂,y₂) = (−5.0 cm, 0 cm) in component form

here , q =  12 x 10⁻⁹ C , d = 5+1 = 6  cm

E = 9 X 10⁹ x  12 x 10⁻⁹ /( 6 x 10⁻²)²

=  3   x 10⁴ N/C

It will act towards the origin along + x axis.

E = 3 x 10⁴ i

electric field at the position (x₃,y₃) = (0 cm, 5.0 cm) in component form

distance between point at ( 1 , 0 ) and (x₃,y₃) = (0 cm, 5.0 cm)

d² = (1² + 5²) = 26 cm²

the electric field at the position   (x₃,y₃) = (0 cm, 5.0 cm) ) in component form

here , q =  12 x 10⁻⁹ C , d² =  26  cm

E = 9 X 10⁹ x  12 x 10⁻⁹ / 26 x 10⁻⁴

=  4.15 x 10⁴ N/C

E in vector form

E = - 4.15 x 10⁴ cosθ i +4.15 x 10⁴ sinθ j where cosθ = 1 / √26 ,

sinθ = 5 / √26

E = - 4.15 x 10⁴ / √26 i + 4.15 x 5 X 10⁴ /√26  j

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation: