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Bezzdna [24]
4 years ago
7

Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa

l to zero at this point? If not, then provide an example to prove your answer. If you think that the answer is "yes", please justify it.
Physics
1 answer:
lyudmila [28]4 years ago
8 0

Answer:

  • No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field \vec{E} is related to the gradient of the electric potential V with :

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})

This means that for constant electric potential the electric field must be zero:

V(\vec{r}) = k

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k

\vec{E} (\vec{r}) = -  (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k

\vec{E} (\vec{r}) = -  (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})

\vec{E} (\vec{r}) = -  (0,0,0)

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4

give an electric field of zero at point (0,0,0)

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3 years ago
A 1-m3 tank containing air @ 25 oC & 500 kPa is connected to another tank containing 5 kg of air at 35 oC & 200 kPa thro
VladimirAG [237]

Answer:

Volume of Tank 2, V' = 2.17 m^{3}

Equilibrium Pressure, P_{eq} = 278.82 kPa

Given:

Volume of Tank 1, V = 1 m^{3}

Temperature of Tank 1, T = 25^{\circ}C = 298 K

Pressure of Tank 1, P = 500 kPa

Mass of air in Tank 2, m = 5 kg

Temperature of tank 2, T' = 35^{\circ}C = 303 K

Pressure of Tank 2, P' = 200 kPa

Equilibrium temperature, 20^{\circ}C = 293 K

Solution:

For Tank 1, mass of air in tank can be calculated by:

PV = m'RT

m' = \frac{PV}{RT}

m' = \frac{500\times 1}{0.287\times 298} = 5.85 kg

Also, from the eqn:

PV' = mRT

V' = volume of Tank 2

Thus

V' = \frac{mRT}{P}

V' = \frac{5\times 0.287\times 303}{200} = 2.17 m^{3}

Now,

Total Volume, V'' = V + V' = 1 + 2.17 = 3.17m^{3}

Total air mass, m'' = m + m' = 5 + 5.85 = 10.85 kg

Final equilibrium pressure, P'' is given by:

P_{eq}V'' = m''RT_{eq}

P_{eq} = \frac{m''RT_{eq}}{V''}

P_{eq} = \frac{10.85\times 0.87\times 293}{3.17} = 287.82 kPa

P_{eq} = 287.82 kPa

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3 years ago
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