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Two identical spheres of mass 1 kg are placed 1 $\mathrm{m}$ apart from eachother. Each sphere pulls on the other with a gravitational force, $F_{\mathrm{g}} .$ Ifeach sphere also holds 1 $\mathrm{C}$ of positive charge, then the magnitude ofthe resulting repulsive electric force is(A) $1.82 \times 10^{40} \mathrm{F}_{\mathrm{g}}$(B) $1.35 \times 10^{20} \mathrm{F}_{\mathrm{g}}$(C) $7.42 \times 10^{-21} \mathrm{F}_{\mathrm{g}}$(D) $5.50 \times 10^{-41} \mathrm{F}_{\mathrm{g}}$

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Chapter 1

Practice Test 1

Section 1

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here we have two spheres which exert gravitational force on each other and they also have charged and repel each other with electric force. So gravitational force is given by Newton's Law of gravitation. G. M one M 2 bye. R squared are the separation between them and column force is given by which is a repulsive force between the two charges. That is given by if the electric force In the column 4th and they were given by K. Which is columns constant, Cuban, too, the two charges and the separation between them. Now from this we can find ratio of column forced to gravitational force equal to Cuban Q two by our square divided by G. Everyone I am too by hi square. No separation between them will remain the same for both the forces. So our square cancels out and we are left with Okay, Cuban Q two by G M one M 2. No K has value nine into 10. To the power line. Newton meter square park, column square. And you want you to is one column each, and M one M two is one kg each, and The value of gravitational constant is 6.67 into 10 to the power -11 Newton meter square park A g square. So yeah, we solve this and we find F e by G equal to 1.35 To 10 to the power 20 so the magnitude of electric force is equal to. This is the answer for the electric repulsive force. Hence the correct option is B. This complete resolution. Thank you. Okay.

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