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Anna71 [15]
3 years ago
12

An electric dipole is made of two charges of equal magnitudes and opposite signs. The positive charge, q=1.0 μC, is located at t

he point (x, y, z) = (0.00 cm, 1.0 cm, 0.00 cm), while the negative charge is located at the point (x, y, z) = (0.00 cm, −1.0 cm, 0.00 cm). How much work will be done by an electric field E⃗ =3.0×106 N/C) i^ to bring the dipole to its stable equilibrium position?
Engineering
1 answer:
vfiekz [6]3 years ago
8 0

Answer:

work done by electric field  is 0.06 J

Explanation:

Given data:

Two point charge is + 1\mu C  and -1 \mu C

0+1 charge positioned is (0 cm , 1 cm, 0.00 cm)

-1 charge positioned is (0 cm , -1 cm, 0.00 cm)

E = 3.0\times 10^6 N/C

From above information, the distance between  given two charges d = 2 cm

then d = 0.02m

 work needed is W = q E d

W = 1.0 \times 10^{-6} \times 3.0 \times 10^6 \times 0.02

W = 0.06 J  

Therefore work done by electric field  is 0.06 J

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Expalin the application of diesel cycle in detail.
mars1129 [50]

Explanation:

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4.Constant volume heat rejection

In general compression ratio in diesel engine is high as compare to petrol engine.But the efficiency of diesel cycle is less as compare to petrol cycle for same compression ratio.

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5 0
3 years ago
Investigating how slime molds reproduce is an example of applied research.<br> True<br> False
azamat

Answer:

false

Explanation:

3 0
3 years ago
An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

sat vapor

m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

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4 0
2 years ago
Why is logging done during drilling?
Solnce55 [7]

Answer:

Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible

8 0
2 years ago
Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q whil
Karo-lina-s [1.5K]

Answer:

E = \frac{3Q}{2A\epsilon_0}

Explanation:

By Gauss Law for electric field:

E = \frac{\sigma}{2\epsilon_0}

Where \sigma is the charge density Q/A. Since we have 2 parallel  plates with different charge, the electric field at point P in the gap would be the sum of 2 field

E = E_1 + E_2

E = \frac{Q}{2A\epsilon_0} + \frac{2Q}{2A\epsilon_0}

E = \frac{3Q}{2A\epsilon_0}

5 0
3 years ago
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