## Two charges are located in the x x – y y plane. If q 1 = − 2.75 nC q1=−2.75 nC and is located at ( x = 0.00 m , y = 0.600 m ) (x=0.00 m,y=0.

Two charges are located in the x x – y y plane. If q 1 = − 2.75 nC q1=−2.75 nC and is located at ( x = 0.00 m , y = 0.600 m ) (x=0.00 m,y=0.600 m) , and the second charge has magnitude of q 2 = 3.40 nC q2=3.40 nC and is located at ( x = 1.30 m , y = 0.400 m ) (x=1.30 m,y=0.400 m) , calculate the x x and y y components, E x Ex and E y Ey , of the electric field, → E E→ , in component form at the origin, ( 0 , 0 ) (0,0) . The Coulomb force constant is 1 / ( 4 π ϵ 0 ) = 8.99 × 10 9 N ⋅ m 2 /C 2 1/(4πϵ0)=8.99×109 N⋅m2/C2 .

## Answers ( )

Answer:Explanation:In this case we have to work with vectors. Firs of all we have to compute the angles between x axis and the r vector (which points the charges):

the electric field has two components Ex and Ey. By considering the sign of the charges we obtain that:

Hence, by replacing E1 and E2 we obtain:

hope this helps!!