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the electric field between the plates of an ideal parallel plate capacitor has magnitude 2000 v/m what is the change in potential energy of the system as a 2c charge is moved from a to b?

User Sirrah
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1 Answer

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Final answer:

The change in potential energy of the system as a 2C charge is moved from point A to point B can be calculated using the formula: ΔPE = q * ΔV. In this case, the charge q is 2C and the potential difference ΔV is the voltage between the plates. Therefore, the change in potential energy of the system is 40 Joules.

Step-by-step explanation:

The change in potential energy of the system as a 2C charge is moved from point A to point B can be calculated using the formula:

ΔPE = q * ΔV

Where:

  • ΔPE is the change in potential energy
  • q is the charge being moved
  • ΔV is the change in potential difference between the two points

In this case, the charge q is 2C and the potential difference ΔV is the voltage between the plates. Since the electric field between the plates is given as 2000 V/m, we can assume that the plates are parallel and uniform and use the formula:

ΔV = E * d

Where:

  • ΔV is the change in potential difference
  • E is the electric field strength
  • d is the distance between the plates

Substituting the given values into the equations:

  • ΔV = 2000 V/m * d = 2000 V/m * 0.01 m = 20 V
  • ΔPE = 2C * 20 V = 40 J

Therefore, the change in potential energy of the system is 40 Joules.

User Markus Kasten
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