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xxTIMURxx [149]
3 years ago
15

How do charges move through an insulated wire connected across a battery? Use the terms potential difference, current, conductor

, and insulator in answer.
Physics
2 answers:
mario62 [17]3 years ago
5 0
The battery has both a positive side and a negative side. The potential difference will be the potential of the positive end of the battery minus the potential of the negative end, which will result in positive potential, or power. This potential is what gives the electrons the ability to flow from the positive end to the negative end, or to make a current. While the outside of the insulated wire is the insulator, the inside of the wire has copper, which is the conductor of the current. So, this current will go through the copper of the wire (and not the insulator) from the positive side to the negative. 
kirill115 [55]3 years ago
5 0

Answer:

Battery is the source of electrical energy which will help the charge to flow through a medium.

Battery is a constant source of potential difference which will create constant potential difference across the ends of a conductor where it is connected.

When a charge is passed through the constant potential difference then the change in electrostatic potential energy will give kinetic energy to the charge.

It is given as

q\Delta V = \frac{1}{2}mv^2

so here charge will continuously flow through the conductor across which potential difference is applied by the battery. This constant motion or flow of charge is known as electric current.

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Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10
11111nata11111 [884]

Answer:

1.62\times 10^{-8}\ \text{s}

Explanation:

\epsilon_0 = Vacuum permittivity = 8.854\times 10^{-12}\ \text{F/m}

A = Area = 10\times 2\times 10^{-4}\ \text{m}^2

d = Distance between plates = 1 mm

V_c = Changed voltage = 60 V

V = Initial voltage = 100 V

R = Resistance = 1000\ \Omega

Capacitance is given by

C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}

We have the relation

V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}

The time taken for the potential difference to reach the required level is 1.62\times 10^{-8}\ \text{s}.

5 0
3 years ago
0.042 kg … ________________grams
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Answer:

42 grams

Explanation:

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Answer: 6N/m

Explanation:

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