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To see why charge density and electric field are larger at the sharp comers of a conductor, consider two metal spheres of radii \(r_{1}=R\) and \(r_{2}=2 R,\) both charged to the same potential \(V_{0}\) a. What is the ratio \(\eta_{1} / \eta_{2}\) of their surface charge densities? b. What is the ratio \(E_{1} / E_{2}\) of the electric field strengths at their surfaces?

Short Answer

Expert verified
The ratio of their surface charge densities is 2, and the ratio of the electric field strengths at their surfaces is also 2. They are both larger at the smaller sphere.

Step by step solution

01

Determine charge densities

The surface charge density of a sphere is given by \(\eta = Q / 4\pi r^{2}\), where Q is the charge on the sphere and r is the radius of the sphere. Since both spheres are charged to the same potential \(V_{0}\), the charge on each sphere is \(Q = C V = 4\pi\epsilon_{0} r V_{0}\), where \(\epsilon_{0}\) is the permittivity of free space. We can substitute this into the equation for \(\eta\). We now have \(\eta = 4\pi\epsilon_{0} r V_{0} / 4\pi r^{2} = \epsilon_{0} V_{0} / r\). For the spheres with radii \(r_{1} = R\) and \(r_{2} = 2R\), we have \(\eta_{1} = \epsilon_{0} V_{0} / R\) and \(\eta_{2} = \epsilon_{0} V_{0} / 2R\). Thus, \(\eta_{1} / \eta_{2} = 2\). The charge density of the first sphere is twice that of the second, larger sphere.
02

Determine electric field strengths

The electric field E outside of a spherical conductor of radius r and charge Q is given by \(E = Q / 4\pi\epsilon_{0} r^{2}\). Again, since both spheres are charged to the same potential \(V_{0}\), the charge on each sphere is \(Q = C V = 4\pi\epsilon_{0} r V_{0}\). Substituting this back into the equation for E, we get \(E = 4\pi\epsilon_{0} r V_{0} / 4\pi\epsilon_{0} r^{2} = V_{0} / r\). For the spheres with radii \(r_{1} = R\) and \(r_{2} = 2R\), we have \(E_{1} = V_{0} / R\) and \(E_{2} = V_{0} / 2R\). Thus, \(E_{1} / E_{2} = 2\). The electric field strength of the first sphere is twice that of the second, larger sphere.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Charge Density
In electrostatics, charge density refers to the amount of electric charge per unit area on the surface of a conductor. When dealing with spheres, the surface charge density \( \eta \) is calculated by dividing the charge \( Q \) by the surface area of the sphere. The formula becomes \( \eta = \frac{Q}{4\pi r^2} \), where \( r \) is the radius of the sphere.

Interestingly, when two spheres are charged to the same electric potential \( V_0 \), their charge distribution behaves differently depending on their size. For spheres with radii \( r_1 = R \) and \( r_2 = 2R \), the charge densities are \( \eta_1 = \frac{\epsilon_0 V_0}{R} \) and \( \eta_2 = \frac{\epsilon_0 V_0}{2R} \).

As seen in the solution, the ratio \( \frac{\eta_1}{\eta_2} = 2 \). This highlights how smaller spheres pack more charge density than larger ones when both are at the same potential.
Electric Field
The electric field describes the force experienced by a charged particle in space, and it is fundamental to understanding electrostatics. The electric field \( E \) produced by a spherical conductor can be derived from Gauss's law. For a sphere charged to a specific potential \( V_0 \), the formula simplifies to \( E = \frac{V_0}{r} \), with \( r \) representing the radius.

For our example with radii \( r_1 = R \) and \( r_2 = 2R \), the electric fields are \( E_1 = \frac{V_0}{R} \) and \( E_2 = \frac{V_0}{2R} \).

This leads to the ratio \( \frac{E_1}{E_2} = 2 \), illustrating how the electric field is more intense on smaller spheres, emphasizing how charge distribution affects electric fields.
Conductors
Conductors are materials that allow electric charges to move freely across their surfaces. This movement creates uniform electric fields within the conductor and ensures that excess charges reside on the surface.

In the example, the metal spheres are conductors, meaning they distribute charge evenly across their surfaces when at equilibrium. This distribution is vital for understanding how charge density and electric field strength vary with the sphere's size. Smaller regions, like sharp corners, can have higher localized charge densities, a concept known as "charge accumulation."

This property is why sharper areas on conductors can exhibit higher electric fields.
Electric Potential
Electric potential, often denoted as \( V \), measures the electric potential energy per unit charge at a certain point in the space. It reflects how much work is needed to move a charge to that point. Importantly, all points on a conductor's surface are at the same potential.

For the spheres in our exercise, maintaining the same potential \( V_0 \) ensures that the electric fields and charge densities adjust according to size. The larger sphere, with radius \( 2R \), has a diminished surface charge density and electric field compared to the smaller sphere, due to its larger radius spreading out the charge.

This equilibrium condition underpins many electrostatic phenomena, ensuring constant potential leads to variable charge distributions depending on geometry.

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Most popular questions from this chapter

You need a capacitance of \(50 \mu \mathrm{F}\), but you don't happen to have a \(50 \mu \mathrm{F}\) capacitor. You do have a \(30 \mu \mathrm{F}\) capacitor. What additional capacitor do you need to produce a total capacitance of \(50 \mu\) F? Should you join the two capacitors in parallel or in series?

An infinitely long cylinder of radius \(R\) has linear charge density \(\lambda\). The potential on the surface of the cylinder is \(V_{0},\) and the electric field outside the cylinder is \(E_{r}=\lambda / 2 \pi \epsilon_{0} r .\) Find the potential relative to the surface at a point that is distance \(r\) from the axis, assuming \(r>R\).

Metal sphere 1 has a positive charge of \(6.0 \mathrm{nC}\). Metal sphere 2\. which is twice the diameter of sphere 1 , is initially uncharged. The spheres are then connected together by a long, thin metal wire. What are the final charges on each sphere?

The electric potential in a region of space is \(V=100\left(x^{2}-\right.\) \(y^{2}\) ) \(V,\) where \(x\) and \(y\) are in meters. a. Draw a contour map of the potential, showing and labeling the \(-400 \mathrm{V},-100 \mathrm{V}, 0 \mathrm{V},+100 \mathrm{V},\) and \(+400 \mathrm{V}\) equipotential surfaces. b. Find an expression for the electric field \(\vec{E}\) at position \((x, y)\) c. Draw the electric field lines on your diagram of part a.

What is the potential difference between \(x_{i}=10 \mathrm{cm}\) and \(x_{f}=\) \(30 \mathrm{cm}\) in the uniform electric field \(E_{x}=1000 \mathrm{V} / \mathrm{m} ?\)

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