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Phantasy [73]
4 years ago
13

A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distancer1

(r1< R) from the center of the sphere, the electric field has magnitude E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R, the magnitude of the electric field at the same distancer1 from the center would be equal to ___________.
Physics
1 answer:
Jobisdone [24]4 years ago
4 0

Answer:

The magnitude of the electric field in the second case will be \frac{1}{8} of the electric field in the first case

Explanation:

The electric field at a distance r_1 < R from the center of the sphere is given by

the formula E_1  = \frac{kQr_1}{R^3}  

where R is the radius of the sphere, and, Q is the charge uniformly distributed on the sphere.

It is given that the same charge Q is distributed uniformly throughout a sphere of radius 2R and we have to find the electric field at same distance r_1 from the center.

In the second case the electric field will be given by

E_2 = \frac{kQr_1}{(2R)^3}  = \frac{kQr_1}{8R^3}  =\frac{1}{8}  \frac{kQr_1}{R^3} = \frac{1}{8} E_1

Therefore the magnitude of the electric field in the second case will be \frac{1}{8} of the electric field in the first case.

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The force required to move a chair 4 meters is 3 newton’s. what amount of work is done
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4 years ago
Two strings on a musical instrument are tuned to play at 196 hz (g) and 523 hz (c). (a) what are the first two overtones for eac
Tems11 [23]
(a) first two overtones for each string:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
f_n = n f_1
where f1 is the fundamental frequency.

So, the first overtone (2nd harmonic) of the string is
f_2 = 2 f_1 = 2 \cdot 196 Hz = 392 Hz
while the second overtone (3rd harmonic) is
f_3 = 3 f_1 = 3 \cdot 196 Hz = 588 Hz

Similarly, for the second string with fundamental frequency f_1 = 523 Hz, the first overtone is
f_2 = 2 f_1 = 2 \cdot 523 Hz = 1046 Hz
and the second overtone is
f_3 = 3 f_1 = 3 \cdot 523 Hz = 1569 Hz

(b) The fundamental frequency of a string is given by
f=  \frac{1}{2L}  \sqrt{ \frac{T}{\mu} }
where L is the string length, T the tension, and \mu = m/L is the mass per unit of length. This part  of the problem says that the tension T and the length L of the string are the same, while the masses are different (let's calle them m_{196}, the mass of the string of frequency 196 Hz, and m_{523}, the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:
\frac{523 Hz}{196 Hz} =  \frac{ \frac{1}{2L}  \sqrt{ \frac{T}{m_{523}/L} } }{\frac{1}{2L}  \sqrt{ \frac{T}{m_{196}/L} }}
and since L and T simplify in the equation, we can find the ratio between the two masses:
\frac{m_{196}}{m_{523}}=( \frac{523 Hz}{196 Hz} )^2 = 7.1

(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
\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L_{523}} \sqrt{ \frac{T}{\mu} } }{\frac{1}{2L_{196}} \sqrt{ \frac{T}{\mu} }} 
And since T and \mu simplify, we get the ratio between the two lengths:
\frac{L_{196}}{L_{523}}= \frac{523 Hz}{196 Hz}=2.67

(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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