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Ch 23: Electric Potential
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 23, Problem 36c

Two large, parallel conducting plates carrying op­posite charges of equal magnitude are separated by 2.202.20 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?

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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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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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Related Practice
Textbook Question

A very long insulating cylinder of charge of radius 2.502.50 cm carries a uniform linear density of 15.015.0 nC/m. If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads 175175 V?

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Textbook Question

Two large, parallel conducting plates carrying op­posite charges of equal magnitude are separated by 2.202.20 cm. If the surface charge density for each plate has magnitude 47.047.0 nC/m2, what is the magnitude of EE in the region between the plates?

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Textbook Question

Two large, parallel conducting plates carrying op­posite charges of equal magnitude are separated by 2.202.20 cm. What is the potential difference between the two plates?

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Textbook Question

Certain sharks can detect an electric field as weak as 1.01.0 μμV/m. To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.51.5­V AA battery across these plates, how far apart would the plates have to be?

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Textbook Question

How much excess charge must be placed on a copper sphere 25.025.0 cm in diameter so that the potential of its center, rela­tive to infinity, is 3.753.75 kV? What is the potential of the sphere's surface relative to infinity?

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Textbook Question

A metal sphere with radius ra r_a is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rbr_b. There is charge +q+q on the inner sphere and charge q-q on the outer spherical shell. Calculate the potential V(r)V(r) for (i) r<rar < r_a; (ii) ra<r<rbr_a < r < r_b; (iii) r>rbr > r_b. (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take VV to be zero when rr is infinite.

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