A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.0 cm. What is the magnitude of the electric force on a third charge (+2.5 µC) placed on the y axis at y = 3.0 cm?

The magnitude of electric force is  

Explanation:

Coulomb's Law:

The force of attraction or repletion is

   [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force  be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore,

we know,

Total force

                              [ r=5]

                            N

The magnitude of electric force is  

the answer is false

explanation:

your answer is, orbit around the nucleus