Answer:
Capacitive Reactance is 4 times of resistance
Solution:
As per the question:
R =
where
R = resistance
f = fixed frequency
Now,
For a parallel plate capacitor, capacitance, C:
where
x = separation between the parallel plates
Thus
C ∝
Now, if the distance reduces to one-third:
Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.
Also,
Also,
Z ∝ I
Therefore,
Solving the above eqn:
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.
