A voltmeter measures the potential difference between two large parallel plates to be 97.0 volts. The plates are 5.0 cm apart. What is the magnitude of the electric field strength between the plates?

Question

Group of answer choices

4,850 N/C

485 N/C

1,940 N/C

4.85 N/C

Answer 1

To find the magnitude of the electric field strength ($E$) between two parallel plates, you can use the formula:

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

where:

• $V$ is the potential difference (voltage) between the plates,
• $d$ is the distance between the plates.

In this case:

• $V=97.0$ volts,
• $d=5.0$ cm = $0.05$ meters (since we need to convert centimeters to meters).

Now, plug the values into the formula:

$$E=\frac{97.0,\text{V}}{0.05,\text{m}}=1940,\text{N/C}$$

Therefore, the magnitude of the electric field strength between the plates is:

1,940 N/C.