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Mashutka [201]
3 years ago
13

Why do electric field lines explain why like charges repel and opposite charges attract?

Physics
1 answer:
vivado [14]3 years ago
5 0

It can be explained as follows: consider the field produced by a positive charge. If we place a positive test charge in this a field, then this charge would move away from the central charge (because like charges repel), while if we place a negative test charge in this field, this charge would move towards the central charge (because opposite charges repel)

Explanation:

Electric fields are vector fields, and they are represented using field lines.

The field lines give indications on both the magnitude and the direction of the electric field. In fact:

  • The magnitude of the field can be inferred from the spacing between the lines: the closer the lines are, the stronger the field, while for a weaker field the lines are more spread apart
  • The direction of the field is given by the direction of the field lines

In particular, by convention the direction of the field lines represent the direction of the force that a positive test charge would feel when immersed in that field: this means that a positive test charge would accelerate in the direction of the field lines, while a negative test charge would accelerate in the direction opposite to the field lines.

This is in agreement with the fact that like charges repel and opposite charges attract. In fact, the lines of the electric field produced by a single-point positive charge point away from the positive charge: if we place a positive test charge in this field, then this charge would move away from the central charge (because like charges repel), while if we place a negative test charge in this field, this charge would move towards the central charge (because opposite charges repel).

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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4 years ago
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An electric fan is running on HIGH. After the LOW button is pressed, the angular speed of the fan decreases to 84.7 rad/s in 2.9
ale4655 [162]

The initial angular speed of the fan will be 55.0 rad/sec. The angular speed of the fan decreases to 84.7 rad/s in 2.96 s.

<h3>What is angular acceleration?</h3>

Angular acceleration is defined as the pace of change of angular velocity with reference to time.

Given data;

Final angular speed,\rm \omega_f = 84.7 rad/s

Initial angular speed, \rm \omega_i = ?

Time period,t= 2.96 s

Angular deceleration = 47.2 rad/s²

\rm \alpha =\frac{\omega_f-\omega_i}{t} \\\\\rm 47.2 =\frac{84.7-\omega_i}{2.96} \\\\ \omega_i = 55.0  \ rad/sec

Hence the initial angular speed of the fan will be 55.0 rad/sec.

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7 0
2 years ago
If, in general, r were calculated as r =v/i, which circuit arrangement in part a of the experiment would have the smallest error
ladessa [460]

Concept: According to Ohm's Law, the flow of electric current through a conductor is directly proportional to the potential difference across it, provided physical conditions (like temperature, pressure, volume etc.) remains same.

v = ir

or, r = v / i

Here, current (i) is measured by Ammeter which should be connected in series of any electrical circuit.

voltage (v) is measured by Voltmeter which should be connected parallel to the external resistance (r).

In the given experiment, the first arrangement of the circuit will show the smallest error because the voltmeter is connected exactly parallel to the external resistance.

In the second arrangement, the voltmeter is connected across external resistance (r) and Ammeter (A) and in this case, the voltmeter will not measure the exact potential drop across the external resistance (r). So, there would be more error.

4 0
3 years ago
Two points A=(+16) and B=(-4) At electric potential respectively find:
Damm [24]

Answer:

a. Point A

b. 20 V

c. 100 J

Explanation:

a. Point A is at a higher potential because there is a positive sign in front of its magnitude. Since it is a positive integral value, and has a higher magnitude than point B which is at -4, point A is thus at a higher potential than point B.

b. The potential difference between the two points ΔV = A - B

= +16 V - (-4 V)

= +16 V + 4 V

= + 20 V

c. The work done, W in moving a charge Q across a potential difference ΔV is W = QΔV

So, since Q = 5 C and ΔV = + 20 V  

W = QΔV

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dsp73

Answer:

It would decrease.

Explanation:

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