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sattari [20]
3 years ago
9

18. An electric dipole is completely inside a closed imaginary surface and there are no other charges. True or false: (a) The el

ectric field is zero everywhere on the surface. (b) The electric field is normal to the surface everywhere on the surface. (c) The electric flux through the surface is zero. (d) The electric flux through the surface could be positive or negative. (e) The electric flux through a portion of the surface might not be zero
Physics
1 answer:
Zina [86]3 years ago
6 0

Answer:

Explanation:

We shall apply Gauss's theorem to calculate the electric flux through a closed surface due to charge contained in side

Total charge inside surface = zero  

So total electric flux through closed surface will be zero.  

or electric flux going out = electric flux coming in  

(a) The electric field is zero everywhere on the surface.  True  

(b) The electric field is normal to the surface everywhere on the surface. - false

(c) The electric flux through the surface is zero. True  

(d) The electric flux through the surface could be positive or negative - false

(e) The electric flux through a portion of the surface might not be zero - True

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A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o
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Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

P=\rho \times g \times h

Here

  • P is the pressure which is given as 68 kPa.
  • ρ is the density of the oil whose SG is 0.92. It is calculated as

                                       \rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\

  • g is the gravitational constant whose value is 9.8 m/s^2
  • h is the height which is to be calculated

                                        P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m

So the height of column is 7.54m

a-3

By the relation of volume and density

M=\rho \times V

Here

  • ρ is the density of the oil which is 920 kg/m^3
  • V is the volume of cylinder with diameter 16m calculated as follows

                             V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3

Mass is given as

                             M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

                            \Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\

Now corrected pressure is as

P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa

Finding the value of height for this corrected pressure as

P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m

The original height of column is 5.98m

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