Respuesta :

Answer:

F = 15.47 N

Explanation:

Given that,

Q = 52 µC

q = 10 µC

d = 55 cm = 0.55 m

We need to find the magnitude of the electrostatic force on q. The formula for the electrostatic force is given by :

[tex]F=k\dfrac{q_1q_2}{d^2}\\\\F=9\times 10^9\times \dfrac{52\times 10^{-6}\times 10\times 10^{-6}}{(0.55)^2}\\\\F=15.47\ N[/tex]

So, the magnitude of the electrostatic force is 15.47 N.

Cricetus

The magnitude of electrostatic force will be "15.47 N".

Electrostatic force

According to the question,

Charges, Q = 52 μC

                q = 10 μC

Distance, d = 55 cm, or

                   = 0.55 m  

Constant, k = 9 × 10⁹

We know the relation,

→ Electrostatic force, F = k [tex]\frac{q_1 q_2}{d^2}[/tex]

By substituting the values, we get

                                      = 9 × 10⁹ × [tex]\frac{10\times 10^{-6}}{(0.55)^2}[/tex]

                                      = 15.47 N

Thus the above answer is correct.

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