Chapter 12, Problem 5SP

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?

To determine

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 $-0.3\text{J}$.

Explanation of Solution

Given Info: The charge is $+6\times {10}^{-4}\text{C}$, the electric potential at bottom plate is $500\text{V}$, the electric potential at top plate is $0\text{V}$.

Write the expression to find the charge in potential energy of the charge.

$\Delta PE=q\left(\Delta V\right)$

Here,

$\Delta PE$ is the change in potential energy

$q$  is the charge

$\Delta V$ is the change in electric potential

Re-write the above equation using initial and final electric potential values.

$\Delta PE=q\left(V_2-V_1\right)$

Here,

$V_2$ is the final electric potential

$V_1$ is the initial electric potential value

Substitute $+6\times {10}^{-4}\text{C}$ for $q$, $0\text{V}$ for $V_2$ and $500\text{V}$ for $V_1$ to find the change in potential energy.

$\begin{array}{c}\Delta PE=\left(+6\times {10}^{-4}\text{C}\right)\left(0\text{V}-500\text{V}\right)\\ =-0.3\text{J}\end{array}$

Conclusion:

The change in potential energy of the charge, when the charge is moved from bottom plate to top plate is $-0.3\text{J}$.

To determine

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 $+6\times {10}^{-4}\text{C}$, the electric potential at bottom plate is $500\text{V}$, the electric potential at top plate is $0\text{V}$.

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.

To determine

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 $+6\times {10}^{-4}\text{C}$, the electric potential at bottom plate is $500\text{V}$, the electric potential at top plate is $0\text{V}$.

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.

To determine

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 $3.846\times {10}^4\text{N/C}$.

Explanation of Solution

Given Info: The charge is $+6\times {10}^{-4}\text{C}$, the electric potential at bottom plate is $500\text{V}$, the electric potential at top plate is $0\text{V}$, The distance between the plates is $1.3\text{cm}$.

Write the expression to find the electric field.

$\overrightarrow{E}=\frac{\overrightarrow{V}}{d}$

Here,

$\overrightarrow{E}$ is the electric field

$\overrightarrow{V}$ is the electric potential

d is the distance between the plates

Substitute $500\text{V}$ for $\overrightarrow{V}$ and $1.3\text{cm}$ for d to find the electric field.

$\begin{array}{c}\overrightarrow{E}=\frac{500\text{V}}{1.3\times {10}^{-2}\text{m}}\\ =3.846\times {10}^4\text{N/C}\end{array}$

Conclusion:

The value of magnitude of the electric field between the plates is $3.846\times {10}^4\text{N/C}$.