Find the maximum potential difference between two parallel conducting plates separated by 0.5 cm of air, given the maximum sustainable electric field strength in air to be 3.0 × 106 V/m. Hint The maximum potential difference between them is V.

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**Problem Statement**

Find the maximum potential difference between two parallel conducting plates separated by 0.5 cm of air, given the maximum sustainable electric field strength in air to be \(3.0 \times 10^6 \, \text{V/m}\).

**Solution**

The formula to find the potential difference (\(V\)) is:

\[
V = E \times d
\]

where \(E\) is the electric field strength, and \(d\) is the distance between the plates.

Given:
- \(E = 3.0 \times 10^6 \, \text{V/m}\)
- \(d = 0.5 \, \text{cm} = 0.005 \, \text{m}\)

Calculate \(V\):

\[
V = 3.0 \times 10^6 \, \text{V/m} \times 0.005 \, \text{m} = 15000 \, \text{V}
\]

Thus, the maximum potential difference between the plates is 15,000 V.

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Transcribed Image Text:**Problem Statement** Find the maximum potential difference between two parallel conducting plates separated by 0.5 cm of air, given the maximum sustainable electric field strength in air to be \(3.0 \times 10^6 \, \text{V/m}\). **Solution** The formula to find the potential difference (\(V\)) is: \[ V = E \times d \] where \(E\) is the electric field strength, and \(d\) is the distance between the plates. Given: - \(E = 3.0 \times 10^6 \, \text{V/m}\) - \(d = 0.5 \, \text{cm} = 0.005 \, \text{m}\) Calculate \(V\): \[ V = 3.0 \times 10^6 \, \text{V/m} \times 0.005 \, \text{m} = 15000 \, \text{V} \] Thus, the maximum potential difference between the plates is 15,000 V. **Interface Features** - **Hint Button**: Provides guidance or tips for solving the problem. - **Message Instructor Link**: Allows students to contact the instructor for further assistance. - **Submit Question Button**: Enables users to submit their answers for evaluation.
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