Question

What must be the distance between point charge q1 = 26.0 µC and point charge q2 = -47.0 µC for the electrostatic force between them to have a magnitude of 5.70 N?

Answer 1

Answer:

answer is 1.39m

Explanation:

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Answer 2

Answer:

The distance between the two expenses have to be 0.0695 meters (or 6.95 cm) for the electrostatic pressure between them to have a magnitude of 5.70 N.

Explanation:

From the above question,

They have given :

The electrostatic pressure between two factor fees q1 and q2 is given by means of Coulomb's law:

                F = ok * |q1*q2| / $r^2$

where F is the magnitude of the electrostatic force, ok is Coulomb's consistent (k = 9.0 x 10^9 N*m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.

In this problem, we have:

            q1 = 26.0 µC = 26.0 x $10^{-6$ C

            q2 = -47.0 µC = -47.0 x $10^{-6$ C

              F = 5.70 N

We prefer to locate the distance r between the charges.

First, we want to use the absolute price of the product of the charges, due to the fact that the pressure is constantly fascinating between contrary charges:

  |q1*q2| = |26.0 x $10^{-6$ C * -47.0 x $10^{-6$ C|

              = 1.222 x $10^{-12$ $C^2$

Now we can rearrange Coulomb's law to clear up for r:

            r = sqrt(k * |q1*q2| / F)

Substituting the given values, we get:

             r = sqrt((9.0 x 10^9 N*$m^2$/$C^2$) * (1.222 x $10^{-12$ $C^2$) / 5.70 N)

             r =  0.0695 m

Therefore,

         The distance between the two expenses have to be 0.0695 meters (or 6.95 cm) for the electrostatic pressure between them to have a magnitude of 5.70 N.

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