Answer:
The charge is [tex]q = 3.14 *10^{-4} \ C[/tex]
Explanation:
From the question we are told that
The mass of each ball is [tex]m = 200 \ lb = \frac{200}{2.205} = 90.70 \ kg[/tex]
The distance of separation is [tex]d = 1 \ m[/tex]
Generally the weight of the each ball is mathematically represented as
[tex]W = m * g[/tex]
where g is the acceleration due to gravity with a value [tex]g = 9.8 m/s^2[/tex]
substituting values
[tex]W = 90.70 * 9.8[/tex]
[tex]W = 889 \ N[/tex]
Generally the electrostatic force between this balls is mathematically represented as
[tex]F_e = \frac{k * q_1* q_2 }{d^2}[/tex]
given that the the charges are equal we have
[tex]q_1= q_2 = q[/tex]
So
[tex]F_e = \frac{k * q^2 }{d^2}[/tex]
Now from the question we are told to find the charge when the weight of one ball is equal to the electrostatic force
So we have
[tex]889 = \frac{9*10^9 * q^2}{1^2}[/tex]
=> [tex]q = 3.14 *10^{-4} \ C[/tex]
The magnitude of charge on the balls is [tex]3.14 \times 10^{-4} \;\rm C[/tex].
Given data:
The masses of two lead balls are, m = 200 lb = 200/2.205 = 90.70 kg.
The distance of separation of two balls is, d = 1 m.
First of all we need to obtain the weight of ball. The weight of the ball is expressed as,
W = mg
Here,
g is the gravitational acceleration.
Solving as,
W = 90.70 × 9.8
W = 888.86 N
The expression for the electrostatic force between this balls is mathematically represented as,
[tex]F = \dfrac{k \times q_{1} \times q_{2}}{d^{2}}[/tex]
Since, the charges are equal then,
[tex]q_{1} =q_{2}=q[/tex]
Also, the magnitude of force between the balls is same as the weight of one ball. Then,
F = W
Solving as,
[tex]F =W= \dfrac{(9 \times 10^{9}) \times q^{2}}{1^{2}}\\\\889= \dfrac{(9 \times 10^{9}) \times q^{2}}{1^{2}}\\\\q = 3.14 \times 10^{-4} \;\rm C[/tex]
Thus, we can conclude that the magnitude of charge on the balls is [tex]3.14 \times 10^{-4} \;\rm C[/tex].
Learn more about the Coulomb's law here:
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