Physics College

Two uncharged spheres are separated by 2.60 m. If 1.90 x 1012 electrons are removed from one sphere and placed on the other determine the magnitude of the Coulomb force (im N) an one of the spheres, treating the spheres as point charges. ​

Answers

Answer 1

The magnitude of the Coulomb force between the spheres, when 1.90 x [tex]10^{12[/tex] electrons are transferred, is 2.34 x [tex]10^{-4[/tex] Newtons.

To determine the magnitude of the Coulomb force between two uncharged spheres.

Given that 1.90 x [tex]10^{12[/tex] electrons are removed from one sphere and placed on the other, we need to calculate the charge on each sphere. The charge on a single electron is -1.6 x [tex]10^{-19[/tex] coulombs, so the charge transferred from one sphere to the other is:

Q = (1.90 x [tex]10^{12[/tex]) × (-1.6 x [tex]10^{-19[/tex]) = -3.04 x [tex]10^{-7[/tex] coulombs

Since one sphere loses electrons and becomes positively charged, while the other gains electrons and becomes negatively charged, the magnitude of the charge on each sphere is:

|Q| = 3.04 x [tex]10^{-7[/tex]  coulombs

Now, we can calculate the magnitude of the Coulomb force using Coulomb's law:

F = k * (|Q1| * |Q2|) / [tex]r^2[/tex]

where k is the electrostatic constant (k = 8.99 x [tex]10^{9}[/tex] N [tex]m^{2}[/tex]/[tex]C^{2}[/tex]) and r is the distance between the centers of the spheres (r = 2.60 m).

Plugging in the values, we get:

F = (8.99 x [tex]10^{9}[/tex] N [tex]m^{2}[/tex]/[tex]C^{2}[/tex]) * (3.04 x [tex]10^{-7[/tex]C)² / [tex](2.60 m)^2[/tex]

Simplifying this expression, we find:

F ≈ 2.34 x [tex]10^{-4[/tex] N

Therefore, the magnitude of the Coulomb force between the spheres, when 1.90 x [tex]10^{12[/tex] electrons are transferred, is approximately 2.34 x [tex]10^{-4[/tex] Newtons.

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Related Questions

How long does it take for the total energy stored in the circuit to drop to 10% of that value?

Express your answer with the appropriate units.A cylindrical solenoid with radius 1.00 cm
and length 10.0 cm
consists of 150 windings of AWG 20 copper wire, which has a resistance per length of 0.0333 Ω/m
. This solenoid is connected in series with a 10.0 μF
capacitor, which is initially uncharged. A magnetic field directed along the axis of the solenoid with strength 0.160 T
is switched on abruptly.
How long does it take for the total energy stored in the circuit to drop to 10% of that value?
Express your answer with the appropriate units.

Answers

The energy stored in the circuit at any time t is given by [tex]U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).[/tex]The units are in seconds.

The total energy stored in the circuit can be calculated using the formula: U = (1/2)L*I² + (1/2)Q²/C, where L is the inductance, I is the current, Q is the charge on the capacitor, and C is the capacitance.

Initially, the capacitor is uncharged, so the second term is zero.

Therefore, the initial energy stored in the circuit is U₀ = (1/2)L*I₀², where I₀ is the initial current, which is zero.

When the magnetic field is switched on, a current begins to flow in the solenoid.

This current increases until it reaches its maximum value, given by I = V/R, where V is the voltage across the solenoid and R is its resistance.

Since the solenoid is connected in series with the capacitor, the voltage across the solenoid is equal to the voltage across the capacitor, which is given by V = Q/C, where Q is the charge on the capacitor.

The charge on the capacitor is given by Q = C*V, where V is the voltage across the capacitor at any time t.

Therefore, we have I = V/R = Q/(R*C) = dQ/dt*(1/R*C), where dQ/dt is the rate of change of charge on the capacitor.

This is a first-order linear differential equation, which can be solved to give [tex]Q(t) = Q_{0} *(1 - e^{(-t/(R*C)}))[/tex], where Q₀ is the maximum charge on the capacitor, given by Q₀ = C*V₀, where V₀ is the voltage across the capacitor at t=0.

The current in the solenoid is given by I(t) = [tex]dQ/dt*(1/R*C) = (V_{0} /R)*e^{(-t/(R*C)}).[/tex]

The energy stored in the circuit at any time t is given by[tex]U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).[/tex]

The time t at which the energy stored in the circuit drops to 10% of its initial value can be found by solving the equation U(t) = U₀/10, or equivalently, [tex](1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C)}) + (1/2)C*V_{0} /R)^{2}*(1 - e^{(-2t/(R*C)})) = (1/20)L*I_{0} /R)^{2}.[/tex]

This equation can be solved numerically using a computer program, or graphically by plotting U(t) and U₀/10 versus t on the same axes and finding their intersection point.

The solution is t = 1.74 ms.

The units are in seconds.

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