Two parallel metal plates 2.0 cm apart, are connected to a 200 V battery
A proton with a positive charge 1.6 x 10-19 cis located between the
plates. The electric field intensity between the plates is
Answers
Answer:
As shown below, when two parallel plates are connected across a battery, the plates become charged and an electric field is established between them.
In this diagram, the battery is represented by the symbol where the longer line is the battery's positive terminal and the shorter line is its negative terminal. Batteries are rated by to their volts where 1 volt is defined as 1 joule per coulomb or 1 V = 1 J/C. This tells us that a 1.5 V battery can "energize" 1 µC of charge by 1.5 µJ. The difference between a 1.5-V AAA battery and a 1.5-V D-cell battery lies in their power. A D-cell can supply its energy for a longer amount of time.
Recall that the direction of an electric field is defined as the direction that a positive test charge would move. So in this case, the electric field would point from the positive plate to the negative plate. Since the field lines are parallel to each other, this type of electric field is uniform and has a magnitude which can be calculated with the equation E = V/d where V represents the voltage supplied by the battery and d is the distance between the plates.
It should now be noted that there are two units in which the electric field strength, E, can be measured: either in V/m, or equivalently, in N/C. That is, expressing the electric field strength as E = 10 V/m is equivalent to stating that the electric field strength equals 10 N/C.
V/d = (J/C)/m
[(nt m)/C]/m
nt/C
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."