All categories

3 years ago

Which affects the electrostatic charge more, charge or distance?

Physics

2 answers:

denpristay [2]3 years ago

8 0

Answer:

Charge

Explanation:

makvit [3.9K]3 years ago

3 0

Answer:

The electric force decrease as the distance increase and vice versa

As it obeys inverse square law that states ththaforce is inversly proportional to square if distance

so if the distance increased to twice the force will be decreased to quarter.

And that is coloumb's law of electrostatic force:

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

F=k(Q1.Q2)/r2

So as you increase the distance between charged objects, Force(attraction/repulsion) weakens by the square of their distance.

As you decrease the distance between charged objects, Force strengthens by the square of their distance.

Inverse relationships are common in nature. In electrostatics, the electrical force between two charged objects is inversely related to the distance of separation between the two objects. Increasing the separation distance between objects decreases the force of attraction or repulsion between the objects. And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. Electrical forces are extremely sensitive to distance. These observations are commonly made during demonstrations and lab experiments. Consider a charged plastic golf tube being brought near a collection of paper bits at rest upon a table. The electrical interaction is so small at large distances that the golf tube does not seem to exert an influence upon the paper bits. Yet if the tube is brought closer, an attractive interaction is observed and the strength is so significant that the paper bits are lifted off the table. In a similar manner, charged balloons are observed to exert their greatest influence upon other charged objects when the separation distance is reduced. Electrostatic force and distance are inversely related.

The pattern between electrostatic force and distance can be further characterized as an inverse square relationship. Careful observations show that the electrostatic force between two point charges varies inversely with the square of the distance of separation between the two charges. That is, the factor by which the electrostatic force is changed is the inverse of the square of the factor by which the separation distance is changed. So if the separation distance is doubled (increased by a factor of 2), then the electrostatic force is decreased by a factor of four (2 raised to the second power). And if the separation distance is tripled (increased by a factor of 3), then the electrostatic force is decreased by a factor of nine (3 raised to the second power). This square effect makes distance of double importance in its impact upon electrostatic

You might be interested in

I NEED HELP ASAP! BRAINIEST TO THE CORRECT ANSWER. HELP ME NOW!

sergij07 [2.7K]

Answer:

$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

V = 12 V
P = 24 W

$\implies \mathsf{24=\frac{12^2}{R} }$

$\implies \mathsf{24R=12^2 }$

$\implies \mathsf{24R=144 }$

$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

Power (P)= 100 W

<em>these lights operate at the usual 240 volts direct from the main electricity supply. Therefore,</em>

V = 240 V

$\implies \mathsf{100=\frac{240^2}{R} }$

<em>R and 100 can interchange places</em>

$\implies \mathsf{R=\frac{240^2}{100} }$

$\implies \mathsf{R=\frac{57600}{100} }$

<u></u>

By Ohm's Law:

$\boxed{\mathsf{Voltage(V)=Current(I) \times Resistance(R)}}$

=> 240 = I × 576

=> I = 0.417 A

I don't know it's resistance,... so sorry

The brightness of the bulb in series is <em><u>less than</u></em> when they're placed individually.

For bulbs in series their resistance gets added to form the equivalent resistance of the two bulbs.

Their resistances are nothing but mere numbers and the sum of two numbers(positive of course) is greater than the numbers.

So, the effective resistance of some bulbs in series <u>is more</u> than the individual resistance.

And

<em>Brightness, i. e., Power</em>

$\boxed{\mathfrak{Power \propto \frac{1}{Resistance} }}$

If resistance increases, Power decreases.

Here, the effective resistance was for sure larger, therefore resistance was increasing, hence power decreased taking brightness along with it.

3 0

3 years ago

Charlie runs at an average speed of 6.5 km/hr. If he runs for 1.5 hours, how far has he traveled?

VikaD [51]

Answer:

9.75 km

Explanation:

Charlie runs 6.5 km/hr

-> Charlie wants to run for 1.5 hours

6.5km + 6.5km/2

= 6.5 km + 3.25km

= 9.75 km

8 0

3 years ago

The temperature that sound moves through the air more quickly

Mademuasel [1]

Sound moves faster in warmer temperature because the particles move faster

7 0

3 years ago

A weight lifter picks up a barbell and 1. lifts it chest high 2. holds it for 30 seconds 3. puts it down slowly (but does not dr

Anna11 [10]

1. lifts it chest high

The force opposing to this action is the force due to gravity. Therefore the work done is:

W1 = m g d

where m is mass of the barbell, g is gravity and d is displacement

2. holds it for 30 seconds

Work is a product of force and displacement, since there is no displacement, therefore work done is zero.

W2 = 0

3. puts it down slowly

If the barbell was dropped, then it would simply be a free fall. But since it was not, so the work done here is also equal to the weight of the barbell times displacement:

W3 = m g d

We can see that W1 = W3, and since W2 = 0, therefore the answer is:

7 0

3 years ago

A boat moves 60 kilometers east from point A to point B. There, it reverses direction and travels another 45 kilometers toward p

Alexxx [7]

Answer:1) the total distance is the sum of the two distances

60 km + 45 km = 105 km

2) The displacement is the net movement, or the difference between the initial position and the final position

Call x the initial position, then the final position is x + [60km - 45km]

And the displacement is x + (60km - 45km) - x =60km -45 km = 15 km

Explanation:

6 0

3 years ago