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7) A force of 500N exists between two identical point charges separated by a dis-
tance of 40 cm the magnitude of the two charges is​

Sagot :

firdausahmed79

Answer:

q=9.43×10^-5C

Explanation:

F=kq^2/r^2

500= 9×10^9 × q^2/ (0.4)^2

500N×0.16m=9×10^9Nm^2C^2 × q^2

80/(9×10^9)= q^2

√(8.8889×10^9) = q

q= 9.43×10^-5C

since they are identical both charges, q=9.43×10^-5C

The magnitude of the two charges is ​9.43×10^-5 C.

To calculate the magnitude of an electric force, it is necessary to use the following expression:

                                          [tex]F = k \frac{q_{1} \times q_{2} }{d^{2}}[/tex]

Assuming that the constant is:

                                          [tex]k = 9\times 10^{9}[/tex]

We can apply the values ​​in the formula above, obtaining:

                                     [tex]500 = 9 \times 10^{9} \times \frac{q^{2}}{0.4^{2}}[/tex]

                                         [tex]q = \sqrt{0.88\times10^{-8}}[/tex]

                                          [tex]q = 9.43 \times 10^{-5} C[/tex]

So, the magnitude of the two charges is   ​9.43×10^-5 C.

Learn more about electrical force in: brainly.com/question/16888648

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