
Concept explainers
Suppose that the top plate of a parallel-plate capacitor has an electric potential of 0 V and the bottom plate has a potential of 500 V. There is a distance of 1.3 cm between the plates.
a. What is the change in potential energy of a charge of +6 × 10–4 C that is moved from the bottom plate to the top plate?
b. What is the direction of the electrostatic force exerted on this charge when it is between the plates?
c. What is the direction of the electric field between the plates?
d. What is the magnitude of the electric field between the plates?
(a)

The change in potential energy of the charge, when the charge is moved from bottom plate to top plate.
Answer to Problem 5SP
The change in potential energy of the charge, when the charge is moved from bottom plate to top plate is
Explanation of Solution
Given Info: The charge is
Write the expression to find the charge in potential energy of the charge.
Here,
Re-write the above equation using initial and final electric potential values.
Here,
Substitute
Conclusion:
The change in potential energy of the charge, when the charge is moved from bottom plate to top plate is
(b)

The direction of the electrostatic force on the charge when it is in between the plates.
Answer to Problem 5SP
The direction of the electrostatic force on the charge when it is between the plates is towards the top plate.
Explanation of Solution
Given Info: The charge is
In the parallel plate capacitor, the direction of the electric field will be in the direction form higher potential to the lower potential. Here, the bottom plate is higher potential and the top plate is lower potential.
Since, the charge is a positive charge; the electrostatic force exerted on the charge will be in the same direction of electric field. Thus, the direction of the electrostatic force on the charge when it is between the plates is towards the top plate.
Conclusion:
The direction of the electrostatic force on the charge when it is between the plates is towards the top plate.
(c)

The direction of the electric field between the plates.
Answer to Problem 5SP
The direction of the electric filed will be form bottom plate to top plate.
Explanation of Solution
Given Info: The charge is
In the parallel plate capacitor, the direction of the electric field will be in the direction form higher potential to the lower potential.
Here, the bottom plate is higher potential and the top plate is lower potential. Thus, the direction of the electric filed will be form bottom plate to top plate.
Conclusion:
The direction of the electric filed will be form bottom plate to top plate.
(d)

The value of magnitude of the electric field between the plates.
Answer to Problem 5SP
The value of magnitude of the electric field between the plates is
Explanation of Solution
Given Info: The charge is
Write the expression to find the electric field.
Here,
d is the distance between the plates
Substitute
Conclusion:
The value of magnitude of the electric field between the plates is
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Chapter 12 Solutions
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