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A very long wire carries a uniform linear charge density λ. Using a voltmeter to measure potential difference, you find that when one probe of the meter is placed 2.30 cm from the wire and the other probe is 1.50 cm farther from the wire, the meter reads 550 V . A) What is |λ|?
B) If you now place one probe at 3.30 cm from the wire and the other probe 1.50 cm farther away, will the voltmeter read 550 V , more or less than 550 V ?

Respuesta :

Answer:

Part a)

[tex]\lambda = 7.15 \times 10^{-8} C/m[/tex]

Part b)

[tex]V = 1014.7 Volts[/tex]

So it is more than 550 V

Explanation:

As we know that the electric field due to linear charge density is given as

[tex]E = \frac{2k\lambda}{r}[/tex]

now we know that potential difference between two points is given as

[tex]\Delta V = \int E.dr[/tex]

[tex]\Delta V = \int(\frac{2K\lambda}{r} dr[/tex]

[tex]\Delta V = 2K\lambda ln(\frac{r_2}{r_1})[/tex]

Part a)

Now we know that potential difference between two points at distance r = 2.30 cm and r = 1.50 cm is given as 550 V

[tex]550 = 2K\lambda ln(\frac{2.30}{1.50})[/tex]

[tex]\lambda = 7.15 \times 10^{-8} C/m[/tex]

Part b)

Now we know that potential difference between two points at distance r = 3.30 cm and r = 1.50 cm is given as

[tex]V = 2K\lambda ln(\frac{3.30}{1.50})[/tex]

[tex]V = 2(9\times 10^9)(7.15\times 10^{-8})(ln(\frac{3.30}{1.50}))[/tex]

[tex]V = 1014.7 Volts[/tex]

So it is more than 550 V

The difference of electric potential between two points is called potential difference.

  • a) The value of [tex]\lambda[/tex] is [tex]7.15\times10^{-8}\rm C/m[/tex]
  • b) The when we place one probe at 3.3 and other at 1.5 cm farther away then voltmeter read more than 550 V.

What is potential difference?

The difference of electric potential between two points is called potential difference.

It can be given as,

[tex]V=2 K \lambda \ln\dfrac{r_2}{r_1}\\[/tex]

Here, [tex]\lambda[/tex] is the energy of photon and [tex]r,_2, r_1[/tex] is the distance of the points.

Given information-

The difference between the two points is 2.30 cm.

Potential difference of the two points is 550 V.

  • A) Value of  |λ|.

Put the values in the above equation as,

[tex]550=2\times K \times \lambda \ln\dfrac{2.3}{1.5}\\\lambda=7.15\times10^{-8}\rm C/m[/tex]

  • B) The reading of the voltmeter for the part B-

In the part B given have,

The position of one probe is 3.30 cm.

The position of another probe is 1.50 cm.

[tex]V=2\times(9\times10^9)(7.15\times10^{-8}\ln\dfrac{3.3}{1.5}\\V=1014.7 \rm V[/tex]

Thus the when we place one probe at 3.3 and other at 1.5 cm farther away then voltmeter read more than 550 V.

Hence,

  • a) The value of [tex]\lambda[/tex] is [tex]7.15\times10^{-8}\rm C/m[/tex]
  • b) The when we place one probe at 3.3 and other at 1.5 cm farther away then voltmeter read more than 550 V.

Learn more about the potential difference here;

https://brainly.com/question/9060304