Question

Two large, parallel metal plates carry opposite charges of equal magnitude. They are separated by $\mathbf{45}\mathbf{.}\mathbf{0}$mm, and the potential difference between them is $\mathbf{360}\mathbf{}\mathbf{\text{V}}$.

(a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?

(b) What is the magnitude of the force this field exerts on a particle with charge $\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{\text{nC}}$?

(c) Use the results of part (b) to compute the work done by the field on the particle as it moves from higher-potential plate to the lower.

(d)Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

Short answer

1. The magnitude of the electric field is $\mathit{E}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{\text{N/C}}$.
2. The magnitude of the force is $\mathit{F}\mathbf{=}\mathbf{19}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathbf{\text{N}}$.
3. The work done by the field on the particle is $\mathit{W}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{64}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}\mathbf{\text{J}}$.
4. The work done equals the negative of the change in potential energy.

Step-by-step solution

Formulas used to solve the question

The relation between electric field and the potential difference between two parallel plates can be written as:

$E=\frac{V_{ab}}{d}$ (1)

Force exerted by electric field can be written as:

$\left|F\right|=\left|q\right|E$ (2)

Work done by the field can be written as:

role="math" localid="1659528351098" $$\begin{gathered} W_{a\to b}=\underset{a}{\overset{b}{\int }}F·dl \\ =Fl \end{gathered}$$ (3)

Where $V_{ab}=$the potential difference between two parallel plates, $d=$the separation between the plates, $E=$electric field,$F=$ electric force, $q=$charge, and $W=$work done.

Compute the magnitude of the electric field

(a)

Given: Two large parallel plates a and b carry opposite charges of equal magnitude are separated by a distance$d=45.0\text{mm}=0.045\text{m}$where the potential difference between two plates is $V_{ab}=360\text{V}$.

The relation between electric field and the potential difference between two parallel plates is given by,

$E=\frac{V_{ab}}{d}$

Plug the values, and we get,

$$\begin{gathered} E=\frac{360\text{V}}{0.045\text{m}} \\ =8\times {10}^3\text{V/m} \\ =8\times {10}^3\text{N/C} \end{gathered}$$

Thus, the magnitude of the electric field is$\mathit{E}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{\text{N/C}}$

Calculate the magnitude of the force

(b)

The force exerted by the electric field is given by,

$\left|F\right|=\left|q\right|E$

Plug the values for $\mathit{q}\mathbf{}$and $\mathit{E}$ into the above expression to get the exerted force,

$$\begin{gathered} \left|F\right|=(2.4\times {10}^{-9}\text{C})(8\times {10}^3\text{N/C}) \\ =19.2\times {10}^{-6}\text{N} \end{gathered}$$

Thus, the magnitude of the force is $\mathit{F}\mathbf{=}\mathbf{19}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathbf{\text{N}}$

Determine the work done by the field

(c)

The work done by the field is given as,

$W_{a\to b}=\underset{a}{\overset{b}{\int }}F·dl=Fl$

Plug the values, and we get,

$$\begin{gathered} W_{a\to b}=(19.2\times {10}^{-6}\text{N})(0.045\text{m}) \\ =8.64\times {10}^{-7}\text{J} \end{gathered}$$

Thus, the work done by the field on the particle is$\mathit{W}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{64}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}\mathbf{\text{J}}$

Determine the work done

(d)

The change in potential energy of the charge q is given by,

$\Delta U=U_b-U_a=q\left(V_b-V_a\right)$

Plug the values, and we get,

$$\begin{gathered} \Delta U=--360\left(2.40\times {10}^{-9}\right) \\ =-8.64\times {10}^{-7}\text{J} \end{gathered}$$

If compared to the result in part (c), the relation can be given as:

$W_{a\to b}=-\Delta U$

The work done equals the negative of the change in potential energy.

Thus, the work done equals the negative of the change in potential energy.