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DiKsa [7]
3 years ago
13

If you have charged an electroscope by contact with a positively charged object, describe how you could use it to determine the

charge of other objects. Specifically, what would the leaves of the electroscope do if other charged objects were brought near its knob?
Physics
1 answer:
maxonik [38]3 years ago
7 0

Explanation:

  • When we have a pre-charged electroscope that is charged using a positive object then we observe that the leaves of the electroscope are spread wide in the diverging position due to the like charges of positive nature on the two leaves.
  • As we know that the charges can only travel through the conductors hence the assembly of the electroscope from the contact knob to the leaves is made of conductor. So the charges spread uniformly on this conductor.
  • Now we can use this as a tester of the other charges. When we bring a negatively charged object near the knob of the electroscope then the opposite charges i.e. positive charges accumulate near the knob due to attraction and the leaves of the electroscope go unstrained and get closer naturally.
  • Contrarily when a positive charge is brought near the knob of the electroscope then the the uniformly distributed charges move away from the knob and accumulate at the farthest point and more deviation is observed on the leaves.
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CHECK COMPLETE QUESTION BELOW

inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA.

Part A)What is the resistance RR of the inductor

PART B) what is inductance L of the conductor

Answer:

A)R=1818.18 ohms

B)L=1.0446H

Explanation:

We were given inductor L with resistance R , there is a connection between the battery and the inductor with Emf of 12V, we can see that the circuit is equivalent to a simple RL circuit.

There is current of 4.96mA at 0.8ms, at the end of the connection the current increase to 6.60mA,

.

a)A)What is the resistance RR of the inductor?

The current flowing into RL circuit can be calculated using below expresion

i=ε/R[1-e⁻(R/L)t]

at t=∞ there is maximum current

i(max)= ε/R

Where ε emf of the battery

R is the resistance

R=ε/i(max)

= 12V/(6.60*10⁻³A)

R=1818.18 ohms

Therefore, the resistance R=1818.18 ohms

b)what is inductance L of the conductor?

i(t=0.80ms and 4.96mA

RT/L = ⁻ln[1- 1/t(max)]

Making L subject of formula we have

L=-RT/ln[1-i/i(max)]

If we substitute the values into the above expresion we have

L= -(1818.18 )*(8.0*10⁻⁴)/ln[1-4.96/6.60)]

L=1.0446H

Therefore, the inductor L=1.0446H

3 0
4 years ago
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