To find the spacing between equipotential surfaces that differ by 9.50 V for a uniformly charged infinite sheet with a surface charge density of +6.00 nC/m², we can follow these steps:
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Determine the Electric Field (E): The electric field \( E \) created by an infinite sheet of charge can be calculated using the formula:
\[ E = \frac{\sigma}{2\epsilon_0} \]
where:
- \( \sigma \) is the surface charge density (\(+6.00 , \text{nC/m}^2 = 6.00 \times 10^{-9} , \text{C/m}^2\)),
- \( \epsilon_0 \) is the permittivity of free space (\(8.85 \times 10^{-12} , \text{C}^2/\text{N m}^2\)).
Plugging in the numbers:
\[ E = \frac{6.00 \times 10^{-9}}{2 \times 8.85 \times 10^{-12}} \approx 339 , \text{N/C} \]
The factor of 2 accounts for the contribution of the field from both sides of the sheet. Notice that the direction of the electric field is away from the positively charged sheet.
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Relating Electric Field and Potential Difference: The relationship between the electric field and the potential difference \( V \) over a distance \( d \) is given by:
\[ V = E \cdot d \]
We need to rearrange this equation to find the distance \( d \):
\[ d = \frac{V}{E} \]
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Calculate the distance for a 9.50 V potential difference:
\[ d = \frac{9.50 , \text{V}}{339 , \text{N/C}} \approx 0.02796 , \text{m} \approx 0.028 , \text{m} = 28 , \text{mm} \]
Thus, the spacing between equipotential surfaces that differ by 9.50 V is approximately 0.028 m (or 28 mm).
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