<span>When two electrical charges, of opposite sign and equal
magnitude, are separated by a distance, a dipole is established. The size of a
dipole is measured by its dipole moment</span><span> . Dip</span>ole moment is measured in Debye units, which is equal to the
distance between the charges multiplied by the charge
<span>Dipolemoment 'D' = e x d
where e= charge and d= distance by which charge is separated. </span><span>
In this case e = charge on electron = 1.6 x10^-19 C</span>
D = 1.6 x10^-19 C (
1x10^-10 m) =1.6x10^-29 Cm
Explanation:
It is given that,
Mass of an electron, 
Initial speed of the electron, 
Final speed of the electron, 
Distance, d = 5 cm = 0.05 m
(a) The acceleration of the electron is calculated using the third equation of motion as :
Force exerted on the electron is given by :
(b) Let W is the weight of the electron. It can be calculated as :
Comparison,
Hence, this is the required solution.
The magnitude of the E-field decreases as the square of the distance from the charge, just like gravity.
Location ' x ' is √(2² + 3²) = √13 m from the charge.
Location ' y ' is √ [ (-3)² + (-2)² ] = √13 m from the charge.
The magnitude of the E-field is the same at both locations.
The direction is also the same at both locations ... it points toward the origin.
Answer:
R/l = 0.25925 Ω / m
Explanation:
Ohm's law says that the potential difference is proportional to the product of the resistance by the current
V = I R
R = V / I
In this case, since we have two lengths, we can have two lengths, we can find the resistance for each
L = 5 m
R = 7.70 / 5.47
R = 1,408 Ω
L = 10 m
R = 7.70 / 3.25
R = 2,369 Ω
We can make a direct proportions rule (rule of three) to find the resistance per unit length
For L = 5 m
R/l = 1,408 / 5
R/l = 0.2816 Ω / m
For L = 10 m
R/l = 2,369/10
R/l = 0.2369 Ω / m
We can see that the value is similar that differs from the second decimal place, in this case the value for the longer re wire is more accurate because it has a lower joule effect.
One way also to find the average value
R/l = (0.2816 + 0.2369) / 2
R/l = 0.25925 Ω / m
