Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.
What is Columb's law?
The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.
They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.
The given data in the problem is;
q₁ is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C
F is the repulsive force = 36 mN =36 ×10⁶ N
d is the distance = 0.70 m
The Coulomb force is found as;
[tex]\rm F = \frac{Kq_1q_2}{r^2}\\\\\ \rm 36\times 10^6 = \frac{9 \times 10^9 }{(0.7)^2} \times 4.4 \times 10^{-9} \times q_2\\\\\ q_2 = 8.6241 \times 10^{-19 } \ C[/tex]
Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.
To learn more about Coulomb's law, refer to the link;
brainly.com/question/1616890
#SPJ2
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
