Question

Question: Two large parallel metal plates are 1.5 cmapart and have charges of equal magnitudes but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +5.0 V, what is the electric field in the region between the plates?

Short answer

Answer:

The electric field in the region between the plates is $6.7\times {10}^2\text{V/m}$.

Step-by-step solution

The given data

1. Separation between the two metal plates, r = 1.5 cm
2. Charges are equal in magnitude and opposite in direction.
3. The potential at the negative plate,$V_{left}=0\text{V}$
4. Potential halfway between the plates,$V_{electric}=+5\text{V}$

Understanding the concept of the electric field

Using the given data, we can get the potential difference between the plates. Now, using this value, we can calculate the electric field at a point using the given concept.

Formula:

The electric field at the point due to the potential difference, $E=\frac{\Delta V}{r}$ (i)

Calculation of the electric potential between the plates

Now, the change in voltage using the given data can be calculated as follows:

$$\begin{gathered} \Delta V=2(5.0\text{V}-0.0\text{V)} \\ =10.0\text{V} \end{gathered}$$

Thus, the magnitude of electric field can be calculated using the given data in equation (i) as follows:

$$\begin{gathered} E=\frac{10.0\text{V}}{1.5\times {10}^{-2}\text{m}} \\ =6.7\times {10}^2\text{V/m} \end{gathered}$$

Hence, the value of the potential is $6.7\times {10}^2\text{V/m}$.