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Dahasolnce [82]
3 years ago
6

From the expression for EE obtained in Problem 22.42,22.42, find the expressions for the electric potential V as a function of r

, both inside and outside the cylinder. Let V=0 at the surface of the cylinder. In each case, express your result in terms of the charge per unit length λ of the charge distribution. (b) Graph V and E as functions of rr from r=0 to r=3R.
Physics
1 answer:
Oksanka [162]3 years ago
4 0

Answer:

a) V(r) =  k*λ*r^2/R^2      r =< R

   V(r) =  -2*k*λ*Ln|r/R|  

Explanation:

Given:

- The derived results for Electric Fields are:

             

             r =< R,              E(r) =  2*k*λ*r / R^2

             r > R,                E(r) = 2*k*λ/ r

Find:

-Expressions for the electric potential V as a function of r, both inside and outside the cylinder.

Solution:

- From definition we can establish the relation between E(r) and V(r) as follows:

                                  E(r) = - dV / dr

- We will develop expression for each case as follows:

         Case 1: r =< R      

                                  E(r) =  2*k*λ*r / R^2 = - dV / dr

         Separate variables:

                                            2*k*λ*r . dr / R^2 = -dV

         Integrating both sides:

                                            2*k*λ/R^2 integral(r).dr = -integral(dv)

                                               k*λ*r^2/R^2 | = - ( 0 - V)

         Put limits ( 0-r)                

                                               V(r) =  k*λ*r^2/R^2                      

          Case 2: r >= R      

                                            E(r) =  2*k*λ/ r = - dV / dr

         Separate variables:

                                                      2*k*λ.dr/ r = -dV

         Integrating both sides:

                                            2*k*λ integral(1/r).dr = -integral(dv)

                                               2*k*λ*Ln|r/R| = - ( V - 0)

         Put limits ( R -> r)                

                                              V(r) =  -2*k*λ*Ln|r/R|                      

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Explanation:

Given that,

Capacitor = 30μC

Resistor = 49.0Ω

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Frequency = 60.0 Hz

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Using formula of impedance

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We need to calculate the value of X_{c}

Using formula of X_{c}

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Put the value of X_{c} into the formula of impedance

Z=\sqrt{(49.0)^2+(88.42)^2}

Z=101.08\ \Omega

(a). We need to calculate the rms current in the circuit

Using formula of rms current

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I_{rms}=\dfrac{30.0}{101.08}

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The rms current in the circuit is 0.30 A.

(b). We need to calculate the rms voltage drop across the resistor

Using formula of rms voltage

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V_{rms}=0.30\times49.0

V_{rms}=14.7\ V

The rms voltage drop across the resistor is 14.7 V

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V_{rms}=I_{rms}\times X_{C}

V_{rms}=0.30\times88.42

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QUESTION 10
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The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

\theta =37.01^{\circ}

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

T_{1}sin(\theta)-T_{2}sin(\theta)=0    (1)

Where:

  • T(1) is the tension due to the rope 1
  • T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

T_{1}cos(\theta)+T_{2}cos(\theta)-W=0   (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)=m_{steel}g

T(1) must be equal to 5479 N, so we have:

cos(\theta)=\frac{m_{steel}g}{2T_{1}}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=0.80

Therefore, the maximum angle allowed is θ = 37.01°.

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