Question

Two charges are located in the xx–yy plane. If q1=−4.10 nCq1=−4.10 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.080 m), and the second charge has magnitude of q2=3.60 nCq2=3.60 nC and is located at (x=1.20 m,y=0.600 m)(x=1.20 m,y=0.600 m), calculate the xx and yy components, ExEx and EyEy, of the electric field, E⃗ E→, in component form at the origin, (0,0)(0,0). The Coulomb force constant is 1/(4π????0)=8.99×109 N⋅m2/C21/(4πϵ0)=8.99×109 N⋅m2/C2.

Answer 1

Answer:

Explanation:

Due to first charge , electric field at origin will be oriented towards - ve of y axis.

magnitude

Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j

= - 31.6 j N/C

Due to second charge electric field at origin

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.2²+ .6²

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.8

= 18 N/C

It is making angle θ where

Tanθ = .6 / 1.2

= 26.55°

this field in vector form

= - 18 cos 26.55 i - 18 sin26.55 j

= - 16.10 i - 8.04 j

Total field

= - 16.10 i - 8.04 j + ( - 31.6 j )

= -16.1 i - 39.64 j .

Ex = - 16.1 i

Ey = - 39.64 j .