An air-filled parallel plate capacitor has circular plates with radius r=20.0 cm, separated distance 4.00 mm. The capacitor is c
1 answer:
You might be interested in
Answer 1) The electric field at distance r from the thread is radial and has magnitude
E = λ / (2 π ε° r)
The electric field from the point charge usually is observed to follow coulomb's law:
E = Q / (4 π ε° )
Now, adding the two field vectors:
= {2.5 / (22 π ε° X 0.07 ) ; 0}
Answer 2) = {2.3 / (4 2 π ε°) ( - 7/ (√(84); -12 / (√84))
Adding these two vectors will give the length which is magnitude of the combined field.
The y-component / x-component gives the tangent of the angle with the positive x-axes.
Please refer the graph and the attachment for better understanding.
Answer:
Range,
Explanation:
The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.
Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.
So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops
Therefore {tex}R = MV²/2QE{/tex}
Answer:
57,42 KJ
Explanation:
By a isobaric proces, the expresion for the works in the jpg adjunt. Then:
W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)
By the ideal gases law: PV=RTn
Then, in (1): (remember Pa = Pb)
W = R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)
Since we have 1 Kg air: How much is this in moles?
From bibliography: 28.96 g/mol
Then, in 1 Kg (1000 g) there are:
n = 34,53 mol
Finally, in (2):
W = (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ
