91Ó°ÊÓ

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,$V_{electric}=+5\text{V}$

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, $E=\frac{\mathrm{Δ}V}{r}$ (i)

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

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

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

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

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