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A solid conducting sphere of radius \(5.00 \mathrm{~cm}\) carries a net charge. To find the value of the charge, you measure the potential difference \(V_{A B}=V_{A}-V_{B}\) between point \(A,\) which is \(8.00 \mathrm{~cm}\) from the center of the sphere, and point \(B\), which is a distance \(r\) from the center of the sphere. You repeat these measurements for several values of \(r>8.00 \mathrm{~cm} .\) When you plot your data as \(V_{A B}\) versus \(1 / r,\) the values lie close to a straight line with slope \(-18.0 \mathrm{~V} \cdot \mathrm{m}\). What does your data give for the net charge on the sphere? Is the net charge positive or negative?

Short Answer

Expert verified
The net charge found on the sphere is approximately -2 nC indicating a negative net charge.

Step by step solution

01

Identify relevant formula

Firstly, note that the potential \(V\) at a distance \(r\) from a point charge \(Q\) can be obtained using the formula: \[V = k \frac{Q}{r}\] Where \(k = 8.99 × 10^9 N m^2/C^2\), is Coulomb’s constant.
02

Use linear graph information

Since the graph plots \(V_{AB}\) versus \(\frac{1}{r}\) and shows a straight line, this indicates an inverse relation as described by above formula, which takes the form: \[V = kQ \times \(\frac{1}{r}\)\] Given the slope of given straight line is -18 V.m, hence we get: \[kQ = -18 V.m\]
03

Solve for the net charge on the sphere

With \(k\) known, solve for \(Q\) to obtain: \[Q = -18 V.m \ast \(\frac{1}{k}\)\] Using \(k= 8.99 × 10^9 N m^2/C^2\), calculate \(Q\) to find net charge as: \[Q = -18 V.m \ast \(\frac{1}{8.99 × 10^9 N m^2/C^2}\) = -2.00 × 10^-9 C ≈ -2 nC\]
04

Interpret the sign of charge

The negative sign indicates that the net charge on the sphere is negative, consistent with the negative slope of the graph given in the question.

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Key Concepts

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

Conducting Sphere Charge
Understanding the behavior of charge on a conducting sphere is crucial to fundamental concepts in electrostatics. A conducting sphere, when charged, distributes the charge uniformly across its surface. This is due to the repulsive forces among the charges that cause them to move as far apart as possible, a phenomenon that can be attributed to the movement of free electrons in a conductor.

When a conducting sphere is placed in an electric field, its charges rearrange, resulting in an effect known as electrostatic shielding. Inside the conducting material of the sphere, the electric field is zero, while on the surface of the sphere, the field lines are perpendicular to it, indicating a uniform distribution of charge. If we want to determine the charge on a conducting sphere, measures of electric potential at different points around it can provide us with necessary data, as demonstrated in the exercise.
Coulomb's Law
Coulomb's Law is a fundamental principle in electrostatics introduced by Charles-Augustin de Coulomb. It describes the force between two point charges as directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:
\[ F = k \frac{Q_1 \cdot Q_2}{r^2} \]
Where \(F\) is the electrostatic force, \(Q_1\) and \(Q_2\) are the magnitudes of charges, \(r\) is the distance between the charges, and \(k\) is Coulomb's constant. The law applies to point charges and can also be applied to conductors by considering the charge to be located at the center for spherical conductors. This law forms the basis for many calculations in electrostatics, including evaluating the electric potential due to a charge distribution.
Electric Potential Difference
The electric potential difference, often simply referred to as voltage (\(V\)), is the work done to move a unit charge from one point to another within an electric field. It is measured in Volts (\(V\)) and can be expressed as:
\[ V = \frac{W}{Q} \]
Where \(W\) is the work done in joules (\(J\)) and \(Q\) is the charge in coulombs (\(C\)). In the context of the exercise, voltage is measured at two points (\(V_A\) and \(V_B\)) relative to a charged sphere. The difference between the potentials (\(V_{AB} = V_A - V_B\)) provides information related to the magnitude of the charge on the sphere. By plotting this potential difference against the inverse of the distance from the sphere's center, one can deduce the properties of the sphere’s charge.
Inverse Square Law
The inverse square law is a principle that reflects the relationship between the intensity of an effect (such as gravity, light, sound, and electrostatic force) and the distance from the source of that effect. As per this law, the intensity is inversely proportional to the square of the distance from the source. For electrostatic forces, Coulomb's law represents this concept, where the force between two charges diminishes with the square of the distance between them.

In terms of electric potential, the inverse square relationship is also evident. The potential at a distance (\(r\)) from a point charge (\(Q\)) is inversely proportional to \(r\), as seen in the formula: \[ V = k \frac{Q}{r} \] Therefore, if plotting electric potential versus \(1/r\), one would expect a straight line, which reflects the linear relationship that emerges when considering the inverse of the distance in the equation. This in-depth understanding of the inverse square law helps in analyzing the graph presented in the exercise and determining the charge on the conducting sphere.

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Most popular questions from this chapter

Charge \(Q=5.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=12.0 \mathrm{~cm} .\) A small sphere with charge \(q=+3.00 \mu \mathrm{C}\) and mass \(6.00 \times 10^{-5} \mathrm{~kg}\) is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within \(8.00 \mathrm{~cm}\) of the surface of the large sphere?

An alpha particle with kinetic energy \(9.50 \mathrm{MeV}\) (when far away) collides head-on with a lead nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the lead nucleus remains stationary and may be treated as a point charge. The atomic number of lead is \(82 .\) The alpha particle is a helium nucleus, with atomic number \(2 .\) )

A thin insulating rod is bent into a semicircular arc of radius \(a,\) and a total electric charge \(Q\) is distributed uniformly along the rod. Calculate the potential at the center of curvature of the are if the potential is assumed to be zero at infinity.

Three equal \(1.20 \mu \mathrm{C}\) point charges are placed at the corners of an equilateral triangle with sides \(0.400 \mathrm{~m}\) long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

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

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