Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on the other to be LaTeX: 1.35\times10^{-4}N1.35 × 10 − 4 N when they are 20 cm apart. Accidentally, one the the experimenters causes the balls to collide and then repositions them 20 cm apart . Now the repulsive force is found to be LaTeX: 1.406\times10^{-4}N1.406 × 10 − 4 N. What are the initial charges on the two metal balls?

Answers

Answer 1

Answer:

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=(1)/(4\pi\spsilon_0)(q_1q_2)/(d^2)-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$(1)/(4\pi\spsilon_0)=9*10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35*10^(-4)$ N

So, from equation (i)

$1.35*10^(-4)=9*10^9(q_1q_2)/((0.2)^2)$

$\Rightarrow q_1q_2=6*10^(-16)\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$(q_1+q_2)/(2)$

d=20cm=0.2m, and $F_c=1.406*10^(-4)$ N

So, from equation (i)

$1.406*10^(-4)=9*10^9(\left((q_1+q_2)/(2)\right)^2)/((0.2)^2)$

$\Rightarrow (q_1+q_2)^2=2.50*10^(-15)$

$\Rightarrow q_1+q_2=\pm5* 10^(-8)$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5* 10^(-8)\;\cdots(iii)$

$\Rightarrow q_1=5* 10^(-8)-q_2$

The equation (ii) become:

$(5* 10^(-8)-q_2)q_2=6*10^(-16)$

$\Rightarrow -(q_2)^2+5* 10^(-8)q_2-6*10^(-16)=0$

$\Rightarrow q_2=3*10^(-8), 2*10^(-8)$

From equation (iii)

$q_1=2*10^(-8), 3*10^(-8)$

So, the magnitude of initial charges on both the sphere are $3*10^(-8)$ Coulombs $=0.03 \mu C$ and $2*10^(-8)$ Colombs or $0.02 \mu C$ .

Considerion the nature of charges too,

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

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Answers

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What is force?

Force acting on a system can be described as the effect that can change the state of the body of motion or rest. The SI measurement unit of force is Newton (N) and force can be described as a vector quantity. Force can change the direction or the speed of the moving object.

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Learn more about force, here:

brainly.com/question/13191643

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The correct answer is 12N

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