An electron moves 4.5 m in the direction of an electric field of strength of 325N/C. Determine the change in electrical potentia
1 answer:
The change in electrical potential energy of the electron is given by:
where
e is the electron charge
is the potential difference between the initial and final point of the electron.
The electron moves by a distance d=4.5 m in the electric field of intensity
, so the potential difference between the initial and final location of the electron is
where the negative sign is due to the fact that the electron is moving in the direction of the electric field, so it is moving from a point at higher potential to a point of lower potential, so the must be negative.
Therefore, the change in electrical potential energy of the electron is
where
e is the electron charge
The electron moves by a distance d=4.5 m in the electric field of intensity
where the negative sign is due to the fact that the electron is moving in the direction of the electric field, so it is moving from a point at higher potential to a point of lower potential, so the
Therefore, the change in electrical potential energy of the electron is
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Answer:
V_{a} - V_{b} = 89.3
Explanation:
The electric potential is defined by
= - ∫ E .ds
In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.
V_{b} - V_{a} = - ∫ E ds
We substitute
V_{b} - V_{a} = - ∫ (α + β/ y²) dy
We integrate
V_{b} - V_{a} = - α y + β / y
We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m
V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)
V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33
V_{b} - V_{a} = - 89.3 V
As they ask us the reverse case
V_{b} - V_{a} = - V_{b} - V_{a}
V_{a} - V_{b} = 89.3