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Lyrx [107]
3 years ago
10

An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

.7 μC/m2. A thick, infinite conducting slab, also oriented perpendicular to the x-axis occupiees the region between a = 2.8 cm and b = 4.8 cm. The conducting slab has a net charge per unit area of σ2 = 88 μC/m2. What is Ex(P), the value of the x-component of the electric field at point P, located a distance 6.8 cm from the infinite sheet of charge?
Physics
2 answers:
pishuonlain [190]3 years ago
7 0

Answer:

E_{total}=4.82*10^6N/C

vector with direction equal to the axis X.

Explanation:

We use the Gauss Law and the superposition law in order to solve this problem.

<u>Superposition Law:</u> the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

E_{total}=E_1+E_2

<u>Thanks Gauss Law</u> we know that the electric field of a infinite sheet with density of charge σ is:

E=\sigma/(2\epsilon_o)

Then:

E_{total}=(\sigma_1+\sigma_2)/(2\epsilon_o)=(-2.7*10^{-6}+88*10^{-6})/(2*8.85*10^{-12})=4.82*10^6N/C

This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.

slava [35]3 years ago
6 0

Answer:

The total electric field will be 5*10^5 N/C

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Answer:

Time required by boat 1 for the round trip is less than that of boat 2.

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Explanation:

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time = \frac{48}{48}

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Total time taken by boat 1 is,

Total time by boat 1 = 1 hour + 1 hour = 2 hour

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Total time taken by boat 1 for the round trip is 2 hour.

Case 2: Boat 2

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While going to another end

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2 years ago
Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19
Alisiya [41]

Answer:

The workdone is  W = 9.28 * 10^{3} J

Explanation:

From the question we are told that

   The height of the cylinder is  h = 0.588\ m

   The face Area is  A = 4.19 \ m^2

    The density of the cylinder is \rho  =  0.346 * \rho_w

     Where \rho_w is the density of freshwater which has a constant value

              \rho_w = 1000 kg/m^3

     

Now  

     Let the final height of the device under the water be  =  h_f

      Let  the initial volume underwater be = V_n

     Let the initial height under water be  = h_i

      Let the final volume under water be  = V_f

According to the rule of floatation

        The weight of the cylinder =  Upward thrust

This is mathematically represented as

          \rho_c g V_n = \rho_w gV_f

         \rho_c A h = \rho A h_f

So      \frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}

   =>     \frac{h_f}{h_c}  = 0.346

Now the work done is mathematically represented as  

          W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh

               =   \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.

              = \frac{g A \rho}{2}  [h^2 - h_f^2]

              = \frac{g A \rho}{2} (h^2)  [1  - \frac{h_f^2}{h^2} ]

Substituting values

        W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)

        W = 9.28 * 10^{3} J

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