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3 years ago

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A straight, nonconducting plastic wire 9.50 cm long carries a charge density of 130 nC/m distributed uniformly along its length.

It is lying on a horizontal tabletop.A) Find the magnitude and direction of the electric field this wire produces at a point 4.50 cm directly above its midpoint.B)If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 4.50 cm directly above its center.

1 answer:

Leviafan [203]3 years ago

6 0

Answer:

A) E = 3.70*10^{4} N/C

B) E = 2.281*10^3 N/C

Explanation:

given data:

charge density $\lambda = 130*10^{-9} C/m$

length of wire = 9.50 cm

a) at x = 4.5 m above midpoint, electric field is calculated as

$E = \frac{1}{ 2\pi \epsilon} * \frac{ \lambda}{x\sqrt{(x^2/a^2)+1}}$

x = 4.5 cm

midpoint a = 4.5 cm = 0.0475 m

$E =2{\frac{1}{ 8.99*10^9} * \frac{130*10^{-9} }{0.045\sqrt{(4.5^2/4.75^2)+1}}$

E = 3.70*10^{4} N/C

B) when wire is in circle form

$Q = \lambda * L$

$= 130*10^{-9} *9.5*10^{-2}$

= 1.235*10^{-8} C

Radius of circle

$r = \frac{L}{2\pi}$

$r = \frac{9.5*10^{-2}}{2\pi}$

r = 1.511*10^{-2} m

$E = \frac{1}{ 2\pi \epsilon} * \frac{Qx}{(x^2+r^2)^{3/2}}$

$E =8.99*10^{9} * \frac{1.23*10^{-8}*4.5*10^{-2}}{((4.5*10^{-2})^2+(1.511*10^{-2})^2)^{3/2}}$

E = 2.281*10^3 N/C

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So, the first overtone (2nd harmonic) of the string is
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(c) Now the tension T and the mass per unit of length $\mu$ is the same for the strings, while the lengths are different (let's call them $L_{196}$ and $L_{523}$ ). Let's write again the ratio between the two fundamental frequencies
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(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them $T_{196}$ and $T_{523}$ . Let's write again the ratio of the frequencies:
$\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L} \sqrt{ \frac{T_{523}}{m/L} } }{\frac{1}{2L} \sqrt{ \frac{T_{196}}{m/L} }}$
Now m and L simplify, and we get the ratio between the two tensions:
$\frac{T_{196}}{T_{523}}=( \frac{196 Hz}{523 Hz} )^2=0.14$

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