Answer:
The third particle should be at 0.0743 m from the origin on the negative x-axis.
Explanation:
Let's assume that the third charge is on the negative x-axis. So we have:

We know that the electric field is:

Where:
- k is the Coulomb constant
- q is the charge
- r is the distance from the charge to the point
So, we have:

Let's solve it for r(3).
Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.
I hope it helps you!
Answer:

Explanation:
Let the linear charge density of the charged wire is given as

here we can use Gauss law to find the electric field at a distance r from wire
so here we will assume a Gaussian surface of cylinder shape around the wire
so we have

here we have


so we have

I recycle, would this help the climate?
<u>Answer:</u> The energy released in the given nuclear reaction is 3.526 MeV.
<u>Explanation:</u>
For the given nuclear reaction:

We are given:
Mass of
= 41.962403 u
Mass of
= 41.958618 u
To calculate the mass defect, we use the equation:

Putting values in above equation, we get:

To calculate the energy released, we use the equation:

(Conversion factor:
)

Hence, the energy released in the given nuclear reaction is 3.526 MeV.