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lubasha [3.4K]
3 years ago
12

An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the

magnitude of the electric flux througha rectangular area of 1.65 m2 in the xy-plane. N m2/C
Physics
1 answer:
zzz [600]3 years ago
8 0

Answer:

The magnitude of the electric flux is 3.53\ N-m^2/C

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area A= 1.65 m^2

We need to calculate the flux

Using formula of the magnetic flux

\phi=E\cdot A

\phi = EA\cos\theta

Where,

A = area

E = electric field

Put the value into the formula

\phi=2.35\times1.65\times\cos 25^{\circ}

\phi=2.35\times1.65\times0.91

\phi=3.53\ N-m^2/C

Hence, The magnitude of the electric flux is 3.53\ N-m^2/C

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Answer:

The  fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is  

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Explanation:

From the question we are told that

     The time  constant  \tau  =  3

The potential across the capacitor can be mathematically represented as

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In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now  since capacitance is  constant  at  \tau  =  3

        The  energy stored can be evaluated at as

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       V^2 =  0.903  V_o ^2

Hence the fraction of the energy stored in an initially uncharged capacitor is  

      k  = 0.903

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Explanation:

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