Given two 2.00 μC charges, as shown in the figure, and a positive test charge [tex]q = 1.28 \times 10^{-18} \, \text{C}[/tex] at the origin, what is the electric field at the origin due to the two 2.00 μC charges?

```
2.00 μC y

2.00 μC --------•------- q -------•-------- x

x = -0.800 m 0 x = 0.800 m
```

The electric field at the position of the positive test charge is given by twice of the electric field due to one of the 2.00μC charges, as the two charges contribute equally and in the same direction. The calculation of this field uses Coulomb's law.

Explanation:

This is a physics question regarding electrical fields. The electric field intensity is given by the equation E = k*Q/r² where k is Coulomb's constant (approximately 8.99 x 10^9 N m²/C² ), Q is the charge (in this case, 2.00μC, which is 2.00 x 10^-6 C) and r is the distance (0.800m in this case). Therefore, the electric field from each charge is E = (8.99 x 10^9 N m²/C²)*(2.00x10^-6C)/(0.800m)².

Since there are two charges and they are in opposite directions, we should add them up to find the total electric field at the location of the positive test charge.

The resulting calculation is E_total = 2*E = 2*(8.99 x 10^9 N m²/C²)*(2.00x10^-6C)/(0.800m)².

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