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A thin spherical shell with radius \(R_{1}=3.00 \mathrm{~cm}\) is concentric with a larger thin spherical shell with radius \(R_{2}=5.00 \mathrm{~cm}\). Both shells are made of insulating material. The smaller shell has charge \(q_{1}=+6.00 \mathrm{nC}\) distributed uniformly over its surface, and the larger shell has charge \(q_{2}=-9.00 \mathrm{nC}\) distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the fol- (ii) \(r=4.00 \mathrm{~cm}\) lowing distance from their common center: (i) \(r=0\) (iii) \(r=6.00 \mathrm{~cm} ?\) (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

Short Answer

Expert verified
The electric potential at \(r=0\) is \(1800 V\). At \(r=4.00 cm\) it's \(-360 V\) and at \(r=6.00 cm\) it's \(-360 V\). The potential difference between the inner and outer shell is \(-2160 V\), implying that the inner shell is at a higher potential.

Step by step solution

01

Calculate Electric Potential at r=0

First, calculate the electric potential at the center \(r=0\). Note that the point is inside the inner shell, so only the inner shell contributes to the electric potential. Use the equation for electric potential \(V = kQ/r\), where \(k = 8.99×10^9 N·m²/C²\), \(Q = q_1 = +6.00 nC\) and \(r = R_1 = 3.00 cm\).
02

Calculate Electric Potential at r=4.00 cm

Similarly, calculate the electric potential at \(r=4.00 cm\). This point lies between the two shells, hence both will contribute to the electric potential at the point. Sum the electric potential from both the inner shell (using \(Q = q_1\) and \(r = R_1\)) and the outer shell (using \(Q = q_2\) and \(r = 4.00 cm\)).
03

Calculate Electric Potential at r=6.00 cm

Finally, calculate the electric potential at \(r = 6.00 cm\). This point is outside both shells, hence both will contribute to the electric potential at the point. Sum the electric potential from both the inner shell (using \(Q = q_1\) and \(r = R_1\)) and the outer shell (using \(Q = q_2\) and \(r = R_2\)).
04

Calculate Potential Difference

To find the potential difference between the surfaces of the two shells, subtract the potential of the inner shell from that of the outer shell, \(V_2 - V_1 = (kq_2/R_2) - (kq_1/R_1)\). The shell with the higher potential can be identified by the sign of the result. If the result is positive, the outer shell has a higher potential, and if it's negative, the inner shell has a higher potential.

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Key Concepts

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

Electric Potential
The concept of electric potential is akin to the idea of gravitational potential in mechanics—it provides a measure of potential energy per unit charge at a specific point in an electric field. To put it simply, electric potential tells us how much electric potential energy a unit charge would have at a particular location.

For a single charge, the electric potential (\( V \)) can be calculated using the formula: \( V = kQ/r \), where \( k \) is the Coulomb's constant (\( 8.99 \times 10^9 N \cdot m^2/C^2 \) in the International System of Units), \( Q \) is the charge, and \( r \) is the distance from the charge to the point in question.

In our case, when calculating the electric potential at various distances from a charged spherical shell, the distance (\( r \) in the formula) changes depending on where we measure the potential. Inside the shell (\( r < R \) where \( R \) is the radius of the shell), the electric potential is constant because the electric field inside a uniformly charged shell is zero. At the surface of the shell, we use the shell’s radius for \( r \). Beyond the shell, for points outside, the shell's effect is the same as if all of its charge were concentrated at its center.
Uniform Charge Distribution
Uniform charge distribution is a situation where charge is spread out evenly over the surface of an object. In our scenario, the two spherical shells have charges \( q_1 \) and \( q_2 \) distributed uniformly over their surfaces. This uniformity simplifies the calculation of electric potentials because it allows us to treat the spherical shells as point charges once we're calculating potential outside the radius of the shells.

A key point to understand here is that the electric field inside a uniformly charged spherical shell is zero—this is a consequence of Gauss's law in electrostatics. That's why only the charge on the inner shell affects the electric potential at the center (\( r = 0 \) position), since we are inside the inner shell and all the charge from the outer shell has no net effect.

The uniform nature of the charge distribution means that as you move away from the surface of the inner shell to a point between the two shells, both shells will influence the electric potential—but only their charges enclosed within that given radius matter. This is why at \( r = 4.00 cm \) we consider the full charge of the inner shell in the potential calculation, but treat any part of the outer shell's charge beyond this radius as having no effect at that specific point.
Potential Difference
Potential difference, often referred to as voltage, is essentially the difference in electric potential between two points in space. It is a measure of the work done to move a charge between those two points and is expressed in volts. When discussing the potential difference between two spherical shells, one can think of it as measuring the amount of energy per unit charge required to move a test charge from the surface of one shell to another.

To calculate the potential difference between our two shells, we apply the formula \( V_2 - V_1 \), where \( V_2 \) is the potential of the outer shell and \( V_1 \) is the potential of the inner shell. The calculated potential difference tells us how the electric potential changes between the surfaces of the two shells. If the result of \( V_2 - V_1 \) is positive, it means the outer shell is at a higher electric potential than the inner shell; if it's negative, the inner shell is at a higher potential.

Understanding potential difference is crucial because it drives the motion of charges in electric circuits and fields. Charges naturally move from regions of higher potential to regions of lower potential, following the path that takes the least energy. This is analogous to a ball rolling downhill in response to a difference in gravitational potential.

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Most popular questions from this chapter

A metal sphere with radius \(r_{a}\) is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius \(\eta_{b}\). There is charge \(+q\) on the inner sphere and charge \(-q\) on the outer spherical shell. (a) Calculate the potential \(V(r)\) for (i) \(rn\). (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take \(V\) to be zero when \(r\) is infinite. (b) Show that the potential of the inner sphere with respect to the outer is $$ V_{a b}=\frac{q}{4 \pi \epsilon_{0}}\left(\frac{1}{r_{a}}-\frac{1}{n_{b}}\right) $$ (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude \(E(r)=\frac{V_{a b}}{\left(1 / r_{a}-1 / r_{b}\right)} \frac{1}{r^{2}}\) (d) Use Eq. (23.23) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance \(r\) from the center, where \(r>r_{b} .\) (e) Suppose the charge on the outer sphere is not \(-q\) but a negative charge of different magnitude, say \(-Q .\) Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different

A uniform electric field has magnitude \(E\) and is directed in the negative \(x\) -direction. The potential difference between point \(a\) (at \(x=0.60 \mathrm{~m}\) ) and point \(b\) (at \(x=0.90 \mathrm{~m}\) ) is \(240 \mathrm{~V}\). (a) Which point, \(a\) or \(b,\) is at the higher potential? (b) Calculate the value of \(E\). (c) A negative point charge \(q=-0.200 \mu \mathrm{C}\) is moved from \(b\) to \(a\). Calculate the work done on the point charge by the electric field.

A proton and an alpha particle are released from rest when they are \(0.225 \mathrm{nm}\) apart. The alpha particle (a helium nucleus) has essentially four times the mass and two times the charge of a proton. Find the maximum speed and maximum acceleration of each of these particles. When do these maxima occur: just following the release of the particles or after a very long time?

(a) An electron is to be accelerated from \(3.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\) to a halt?

A very long uniform line of charge with charge per unit length \(\lambda=+5.00 \mu \mathrm{C} / \mathrm{m}\) lies along the \(x\) -axis, with its midpoint at the origin. A very large uniform sheet of charge is parallel to the \(x y\) -plane; the center of the sheet is at \(z=+0.600 \mathrm{~m}\). The sheet has charge per unit area \(\sigma=+8.00 \mu \mathrm{C} / \mathrm{m}^{2}\), and the center of the sheet is at \(x=0\). \(y=0 .\) Point \(A\) is on the \(z\) -axis at \(z=+0.300 \mathrm{~m}\), and point \(B\) is on the \(z\) -axis at \(z=-0.200 \mathrm{~m}\). What is the potential difference \(V_{A B}=V_{A}-V_{B}\) between points \(A\) and \(B ?\) Which point, \(A\) or \(B,\) is at higher potential?

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