Electric potential problem with two plates and different points (i'll elaborate)

In summary: So, in summary, the electric potential at point R is 45 V and the electric potential at point S is 30 V. The speed of the mass at R when it reaches point S is 3.6*10^4 m/s.
  • #1
nadeem0322
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Homework Statement


There are two parallel plates. The plate on the left has an electric potential of +60 Volts. The right plate has an electric potential of 0 Volts. The plates are 8.0mm apart. Between the two plates lies a point P 2.0mm from the left plate with a charge -3.5pC. Point R lies a mass right in the middle. Point S lies 2.0mm from the right plate.

A)
i) What is the electric potential at point R?
ii) What is the electric potential at point S?

Point R lies a mass of 4*10^-19 kg with a charge -2pC.
B) What is the speed of the mass at R when it reaches point S if it has an initial velocity of 3.0*10^4 m/s.

Homework Equations


V= kq/r
u=kqq/r



The Attempt at a Solution



I solved this problem for Ai) using kq/r where r is the distance from R and P and q is charge at P. I then added the voltage of 60 volts to that number to get the total potential. I believe this is where I went wrong but I don't see how the 60 voltage plate could not affect the potential at the point. Any help would be great. Thanks
 
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  • #2
If the potential of the RH plate is zero then the potential at a point x away from the RH plate towards the LH plate is 60V*x/d, x = distance from RH plate to R or S, ignoring the extra charge at P. Now the potential due to the extra charge at P is just the extra work you have to apply to a unit positive test charge to bring it from the RH plate to R or S.

The usual definition of potential as being the work on a unit test charge brought from infinity to the observation point does not apply here. That's because the point of zero potential is the RH plate and not infinity.
 

Related to Electric potential problem with two plates and different points (i'll elaborate)

What is the electric potential problem with two plates and different points?

The electric potential problem with two plates and different points is a common problem in electrostatics where two parallel plates with different charges are placed at certain distances from each other. The goal of the problem is to calculate the electric potential at different points between the two plates.

What are the key variables in this type of problem?

The key variables in this type of problem are the distance between the two plates, the magnitude of charges on each plate, and the location of the points where the electric potential is being calculated.

How do you calculate the electric potential at a specific point between the two plates?

To calculate the electric potential at a specific point, you can use the formula V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge on the plate closest to the point, and r is the distance between the point and the plate. You can also use the formula V = (kQ1/r1) - (kQ2/r2) for points between the two plates.

What is the significance of the distance between the two plates?

The distance between the two plates plays a crucial role in determining the electric potential at different points. The closer the plates are, the stronger the electric field will be, resulting in a higher electric potential. As the distance increases, the electric potential decreases.

What are some real-world applications of this type of problem?

This type of problem can be applied to various real-world scenarios, such as calculating the electric potential between two parallel conductors, determining the voltage in a circuit, and understanding the behavior of charged particles in an electric field.

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