Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.202.202.20 cm. The surface charge density for each plate has magnitude 47.047.0 $$nC/m^2. If $$the separation between the plates is doubled while the surface charge density is kept constant at the given value, what happens to the magnitude of the electric field and to the potential difference?

Ch 23: Electric Potential

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 23: Electric Potential

Problema 36c

Capitolo 23, Problema 36c

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by $2.20$ cm. The surface charge density for each plate has magnitude $47.0$ nC/m^2. If the separation between the plates is doubled while the surface charge density is kept constant at the given value, what happens to the magnitude of the electric field and to the potential difference?

Guida verificata passo dopo passo

First, understand that the electric field between two parallel plates with surface charge density $ \sigma $ is given by $ E = \frac{\sigma}{\varepsilon_0} $, where $ \varepsilon_0 $ is the permittivity of free space. This formula indicates that the electric field depends only on the surface charge density and not on the distance between the plates.

Since the surface charge density $ \sigma $ remains constant, the magnitude of the electric field $ E $ will also remain constant even if the separation between the plates is doubled.

Next, consider the potential difference $ V $ between the plates, which is related to the electric field and the separation $ d $ by the equation $ V = E \cdot d $. This shows that the potential difference is directly proportional to the separation between the plates.

If the separation $ d $ is doubled, the potential difference $ V $ will also double, because $ V = E \cdot 2d $ when $ d $ is doubled.

In summary, when the separation between the plates is doubled, the electric field remains unchanged, but the potential difference doubles due to the increased separation.

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Electric Field Between Parallel Plates

The electric field between two parallel plates is uniform and can be calculated using the formula E = σ/ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space. This field is independent of the distance between the plates, meaning that doubling the separation does not affect the magnitude of the electric field.

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Parallel Plate Capacitors

Potential Difference

The potential difference (V) between two points in an electric field is given by V = E * d, where E is the electric field and d is the distance between the points. When the separation between the plates is doubled, the potential difference also doubles, as it is directly proportional to the distance between the plates.

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Potential Difference Between Two Charges

Surface Charge Density

Surface charge density (σ) is the amount of charge per unit area on a surface. In this scenario, it remains constant even if the separation between the plates changes. It is a crucial factor in determining the electric field between the plates, as the field is directly proportional to the surface charge density.

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Surface Charge Density

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