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3 years ago

Why is electrostatic force able to act at a distance

1 answer:

mote1985 [20]3 years ago

7 0

In Electrostatics the electrical force between Two charged objects is inversely Related to the distance of separation between the two objects .

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Does the planet exert a torque on the meteoroid with respect to the center of mass of the planet? Why or why not?

ra1l [238]

Answer:

yes

Explanation:

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3 years ago

A small metal ball is given a negative charge, then brought near to end a of the rod (figure 1). What happens to end a of the ro

erma4kov [3.2K]

What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.

<h3>Electrostatics</h3>

I have attached the image of the rod.

We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.

This means that their fields will cancel.

Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.

This also applies to a strong conducting rod and therefore it is strongly attracted.

Read more about Electrostatics at; brainly.com/question/18108470

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2 years ago

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Tasya [4]

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3 years ago

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An AC source of maximum voltage V0 = 30 V is connected to a resistor R = 50 Ω, an inductor L = 0.6 H, and a capacitor C = 20 µF.

ivolga24 [154]

Hello!

We can begin by solving for the resonance ANGULAR frequency of the circuit.

For an RCL circuit, the resonance angular frequency is given as:
$\omega_0^2 = \frac{1}{LC}\\\\w_0 = \sqrt{\frac{1}{LC}}$

ω₀ = resonance angular frequency (rad/s)

L = Inductance (0.6 H)
C = Capacitance (20 μF)

Plug in the values and solve.

$\omega_0 = \sqrt{\frac{1}{(0.6)(0.00002)} } = 288.675 \frac{rad}{s}$

For an AC power source, the output is usually expressed as:

$V(t) = V_{max}sin(\omega_0 t})$

So, using the appropriate values and setting the source angular frequency equivalent to the circuit's resonance angular frequency:

$V(t) = 30sin(288.675t)$

To find the maximum charge on the capacitor when the frequency of the source is equivalent to the resonance frequency of the circuit (or the angular frequencies are equal), we can begin by finding the maximum voltage across the capacitor.

To find this, however, we must solve for the maximum current across the circuit by finding the total impedance of the circuit. When the circuit is at resonance, the impedance is equivalent to the resistance of the RESISTOR.

So, solve for the maximum current in the circuit using Ohm's Law:

$i = \frac{V}{R}$

In this instance AT RESONANCE:

$I_{Max} = \frac{V_Max}{R}\\\\I_{Max} = \frac{30}{50} = 0.6 A$

Now, we must solve for the capacitive reactance in order to find the maximum voltage across the capacitor. Using the following equation for capacitive reactance:
$X_c = \frac{1}{\omega C}\\\\X_c = \frac{1}{(288.675)(0.00002)} = 173.205 \Omega$

Now that we found the maximum current and capacitive reactance, we can now solve for the maximum voltage across the capacitor:
$V_{C, max} = X_C I_{Max}\\\\V_{C, max} = 173.205 * 0.6 = 103.923 V$