Respuesta :
Answer:
A. -2.16 * 10^(-5) N
B. 9 * 10^(-7) N
Explanation:
Parameters given:
Distance between their centres, r = 0.3 m
Charge in first sphere, Q1 = 12 * 10^(-9) C
Charge in second sphere, Q2 = -18 * 10^(-9) C
A. Electrostatic force exerted on one sphere by the other is:
F = (k * Q1 * Q2) / r²
F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²
F = -2.16 * 10^(-5) N
B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:
Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))
= - 6 * 10^(-9) C
Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C
Hence the electrostatic force between them is:
F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²
F = 9 * 10^(-7) N
a. [tex]-2.16 * 10^{-5} N[/tex]
b. [tex]9 * 10^{-7} N[/tex]
Given:
Distance between their centers, r = 0.3 m
Charge in first sphere, [tex]q_1 = 12 * 10^{-9} C[/tex]
Charge in second sphere, [tex]q_2 = -18 * 10^{-9} C[/tex]
Let's solve for each part one by one:
a. Electrostatic force exerted on one sphere by the other is:
Electrostatic forces :
They are non-contact forces; they pull or push on objects without touching them.
Equation for representing this force is:
[tex]F = \frac{(k * q_1 * q_2)}{r^2} \\\\F =\frac{(9 * 10^9 * 12 * 10^{-9} * -18 * 10^{-9}) }{0.3^2}\\\\F = -2.16 * 10^{-5} N[/tex]
b. In order to find final charges we need to add the charges of both spheres and divided by two:
[tex]q_1 + q_2 = 12 * 10^{-9} + (-18 * 10^{-9})\\\\q_1 + q_2 = - 6 * 10^{-9} C[/tex]
Thus, the value of each charge will be: [tex]-3 * 10^{-9} C[/tex]
Hence, the electrostatic force between them is:
[tex]F =\frac{[9 * 10^9 * (-3 * 10^{-9}) * (-3 * 10^{-9}] }{0.3^2} \\\\F = 9 * 10^{-7} N[/tex]
Thus, the electric force between the two spheres, [tex]F = 9 * 10^{-7} N[/tex]
Find more information about "Force" here:
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