Two charges are located in the x – y plane. if =−4.10 nc and is located at (x=0.00 m, y=0.600 m) , and the second charge has magnitude of =3.80 nc and is located at (x=1.50 m, y=0.650 m) , calculate the x and y components, and , of the electric field, ⃗ , in component form at the origin, (0,0) . the coulomb force constant is 1/()=8.99×109 n⋅m2/c2 .

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

Given:

The electric field at a point due to a charge at a point distance away is given by

where,

The electric field at the origin due to first charge is given by

is the position vector of the origin with respect to the position of the first charge.

Assuming, are the units vectors along x and y axes respectively.

Using these values,

The electric field at the origin due to the second charge is given by

is the position vector of the origin with respect to the position of the second charge.

Using these values,

The net electric field at the origin due to both the charges is given by

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this