ما
An election is located in the uniform electric
field_established between two parallel plates
whese would the electron EXPERIENCE greatest
force?
Answers
Explanation:
Since the field lines are parallel and the electric field is uniform between two parallel plates, a test charge would experience the same force of attraction or repulsion no matter where it is located in the field. That force is calculated with the equation F = qE where both F and E are vector quantities and q is a scalar quantity.
In the diagram above, the distance between the plates is 0.14 meters and the voltage across the plates is 28V. The magnitude of the UNIFORM electric field between the plates would be
E = − ΔV/d
= − (-28 V) / (0.14 m)
= 200 V/m
If a positive 2 nC charge were to be inserted anywhere between the plates, it would experience a force having a magnitude of
F = qE
= (2 x 10–9 C) (200 N/C)
= 4.0 x 10-7 newtons
towards the negative, or bottom plate, no MATTER where it is placed in the region between the plates.
A volt is a scalar quantity that equals a joule per coulomb. Based on this definition, moving a coulomb of charge across a potential difference of 1 volt would require 1 joule of work. Surfaces that have the same potential, or voltage, are called equipotential surfaces and are presented by dotted lines that are always drawn at right angles to the field lines (the solid vectors). Field lines always point from regions of high potential to regions of low potential. In our diagram, the top plate would be at +28 V and is the "high potential plate" while the bottom plate would be at 0 V and is the "low potential plate."
When analyzing electric fields between parallel plates, the equipotential surfaces between the plates would be equally spaced and parallel to the plates. In the diagram shown, we have drawn in six equipotential surfaces, creating seven subregions between the plates.
The distance from one surface to another would equal 0.14/7 or 0.02 meters. Therefore the potential difference from one equipotential surface to the next would equal
ΔV = − Ed cos θ
In our example θ = 0º since our 2 nC positive charge will be moving in the same direction as the field lines; that is, towards the negative plate.
ΔV = − (200 V/m)(0.02 m)
ΔV = − 4 volts
Numbering successively from the top plate (+28 V) to the bottom plate (0 V), our equipotential surfaces would have voltages of 24 V, 20 V, 16 V, 12 V, 8 V, and 4 V respectively.
In general, notice that
when traveling in the direction of the electric field, ΔV would be negative since θ equals 0º and cos 0º = 1. Also note that traveling along a field line always moves the charge from positions of high potential to those of lower potential.
when traveling against the field lines, ΔV would be positive since θ equals 180º and cos 180º = -1. Also the charge would be moving towards a position of higher potential.
Charges possess electric potential energy (EPE) based on their position in an electric field just like masses possess gravitational potential energy (PE) based on their position in a gravitational field.
PE = mgh
EPE = qV
Summarizing what we have learned so far. Our 2 nC charge, no matter where it is placed in the electric field, will always experience a force of 4.0 x 10-7 newtons towards the bottom, negative plate. That constant force would result in the charge being uniformly accelerated towards the bottom plate. As the charge travels downward along any arbitrary field line, it will move between equipotential surfaces and lower its electric potential energy (EPE) from
2 x 10-9 x (28) = 56 x 10-9 J to
2 x 10-9 x (24) = 48 x 10-9 J to
2 x 10-9 x (20) = 40 x 10-9 J to
2 x 10-9 x (16) = 32 x 10-9 J to …… eventually 0 J
Recall that work done ( Wby the field = Fs cos θ where θ = 0º) by a conservative field lowers a particle's potential energy. In this case, it is lowering the charge's electric potential energy (EPE). But remember that the field is also accelerating the charge and thus increasing its kinetic energy (KE).
Wby the field = − ΔEPE = + ΔKE
The amount of work done on the 2 nC charge as it moves between each set of successive equipotential surfaces equals
Wby the field = Fs cos 0º
= (qE)s
= (2 x 10-9 C)(200 N/C)(0.02 m)
= 8 x 10-9 J
Wby the field = Fs cos 0º
= − ΔEPE
= − q(Vfinal - Vinitial)
= − (2 x 10–9 C)(-4 J/C)
= + 8.0 x 10-9 J
Applying conservation of energy, the electric potential energy lost by the charge will be equal to the kinetic energy it gains. This means that the 2 nC charge would gain 8.0 x 10-9 J of KE as it moves between each successive set of equipotential surfaces towards the negative plate. This is exactly what we would be expecting since the charge is being accelerated towards the negative plate.
Note that NO work would be done if the 2 nC charge were to be moved along an equipotential surface since all points on a surface are at the same potential.
Explanation
No matter where a test charge is positioned in the field, it will feel the same force of attraction or repulsion since the field lines are parallel and the electric field is uniform between two parallel plates. The formula F = qE, in which F and E are both vector values and q is a scalar number, is used to compute that force.
The plates are spaced 0.14 meters apart in the preceding figure, and a voltage of 28V is applied across them. The UNIFORM electric field between the plates would have a magnitude of
E = − ΔV/d
= − (-28 V) / (0.14 m)
E= 200 V/m
Anywhere between the plates would experience a force of magnitude 2 nC if a positive 2 nC charge were put there then,
F = qE
= (2 x 10–9 C) (200 N/C)
= 4.0 x 10-7 newtons
∴ F = 4.0 x 10-7 newtons
There would be a gap of 0.14/7, or 0.02 meters, between two surfaces. In light of this, the potential difference between each equipotential surface would equal
ΔV = − Ed cos θ
Since our 2 nC positive charge will be travelling in the same direction as the field lines, or towards the negative plate, = 0o in our example.
ΔV = − (200 V/m)(0.02 m)
ΔV = − 4 volts
Since equals 0o and cos 0o = 1, while moving in the direction of the electric field, V would be negative. Be aware that moving down a field line constantly shifts the charge from high potential spots to low potential ones.
Since is 180o and cos 180o equals -1, the velocity would be positive while moving against the field lines. Additionally, the charge would be advancing to a position with greater potential.
Similar to how masses have gravitational potential energy (PE) based on their position in a gravitational field, charges also have electric potential energy (EPE) depending on their position in an electric field.
PE = mgh
EPE = qV
The amount of work done on the 2 nC charge as it moves between each set of successive equipotential surfaces equals
Wby the field = Fs cos 0º
= (qE)s
= (2 x 10-9 C)(200 N/C)(0.02 m)
= 8 x 10^-9 J
Wby the field = Fs cos 0º
= − (2 x 10–9 C)(-4 J/C)
= + 8.0 x 10^-9 J
Fields = 8.0 x 10^-9 J
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