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:

[tex]q_1=\pm0.03 \mu C[/tex] and [tex]q_2=\pm0.02 \mu C.[/tex]

Explanation:

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

[tex]F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)[/tex]

Where, [tex]\epsilon_0[/tex] is the permitivity of free space and

[tex]\frac{1}{4\pi\spsilon_0}=9\times10^9[/tex] in SI unit.

Before  dcollision:

Charges on both the sphere are [tex]q_1[/tex] and [tex]q_2[/tex], d=20cm=0.2m, and [tex]F_c=1.35\times10^{-4}[/tex] N

So, from equation (i)

[tex]1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}[/tex]

[tex]\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)[/tex]

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

[tex]\frac{q_1+q_2}{2}[/tex]

d=20cm=0.2m, and [tex]F_c=1.406\times10^{-4}[/tex] N

So, from equation (i)

[tex]1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}[/tex]

[tex]\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}[/tex]

[tex]\Rightarrow q_1+q_2=\pm5\times 10^{-8}[/tex]

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.

[tex]\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)[/tex]

[tex]\Rightarrow q_1=5\times 10^{-8}-q_2[/tex]

The equation (ii) become:

[tex](5\times 10^{-8}-q_2)q_2=6\times10^{-16}[/tex]

[tex]\Rightarrow -(q_2)^2+5\times 10^{-8}q_2-6\times10^{-16}=0[/tex]

[tex]\Rightarrow q_2=3\times10^{-8}, 2\times10^{-8}[/tex]

From equation (iii)

[tex]q_1=2\times10^{-8}, 3\times10^{-8}[/tex]

So, the magnitude of initial charges on both the sphere are [tex]3\times10^{-8}[/tex] Coulombs[tex]=0.03 \mu C[/tex] and [tex]2\times10^{-8}[/tex] Colombs or [tex]0.02 \mu C[/tex].

Considerion the nature of charges too,

[tex]q_1=\pm0.03 \mu C[/tex] and [tex]q_2=\pm0.02 \mu C.[/tex]


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Answers

Answer:

t= 10 hr

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It would take 3 hours and 45 minutes

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Answers

Answer:

C 31500

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Answers

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Electrical networks and two-terminal components can be linked either serially or parallelly. The electrical network that results will have two terminals and be capable of operating in either a series or parallel topology.

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Answers

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Answers

Explanation:

Given that,

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Density=mass/volume

For water,

[tex]d=\dfrac{m}{V}\\\\m=dV\\\\m=1\ g/cm^3\times 100\ cm^3\\\\m=100\ g[/tex]

For ethanol,

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Hence, 125 cm³ of ethanol has the same mass as 100 cm³ of water.

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.20 m/s is the answer

Answer:

.20 m/s

Explanation:

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Answers

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Then,

average velocity = 30/0.5

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= 60

Thus, the correct answer would be - 60 km/hour west.

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Hadley cells

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Formation of Convection Currents

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