Problem 21
Question
A positive charge \(q\) is fixed at the point \(x=0, y=0\) and a negative charge \(-2 q\) is fixed at the point \(x=a, y=0\). (a) Show the positions of the charges in a diagram. (b) Derive an expression for the potential \(V\) at points on the \(x\) -axis as a function of the coordinate \(x .\) Take \(V\) to be zero at an infinite distance from the charges. (c) At which positions on the \(x\) -axis is \(V=0 ?\) (d) Graph \(V\) at points on the \(x\) -axis as a function of \(x\) in the range from \(x=-2 a\) to \(x=+2 a\) (e) What does the answer to part (b) become when \(x>\) a? Explain why this result is obtained.
Step-by-Step Solution
Verified Answer
Diagram has charges at \(x=0\) and \(x=a\); potential \(V(x) = \frac{kq}{|x|} - \frac{2kq}{|x-a|}\); \(V=0\) at \(x=a/3\); potential graph shows symmetry; as \(x>a\), \(V\) approaches zero.
1Step 1: Diagram the Charge Positions
In a coordinate plane, place the positive charge \( q \) at \((0, 0)\). Place a negative charge \(-2q\) at \((a, 0)\). Indicate these points on the x-axis showing the charges at \(x = 0\) and \(x = a\).
2Step 2: Derive Potential Expression on x-axis
The electric potential due to a point charge \(q\) at a distance \(r\) is given by \( V = \frac{kq}{r} \), where \(k\) is Coulomb's constant. At a point \(x\) on the x-axis, the potential \(V\) due to both charges is the sum: \[ V(x) = \frac{kq}{|x|} - \frac{2kq}{|x-a|}. \]
3Step 3: Determine Positions Where \(V = 0\)
To find the points on the x-axis where the potential is zero, set \(V(x) = 0\): \[ \frac{kq}{|x|} = \frac{2kq}{|x-a|}. \] Solving this gives \( |x| = 2|x-a| \), resulting in values: \( x = a/3 \) and \( x = a \). Since \( x = a \) is one of the charge points, the solution of interest is \( x = a/3 \).
4Step 4: Graph Potential Function on x-axis
Create a graph of \( V(x) = \frac{kq}{|x|} - \frac{2kq}{|x-a|} \) from \(x = -2a\) to \(x = 2a\). Note the behavior of the graph showing values tend to zero when \( x \) is large and infinite potential changes near the charge points.
5Step 5: Evaluate Potential for \(x > a\)
For \(x > a\), the distances to the charges are large compared to \(a\). Thus, \( V(x) \approx \frac{-qk}{x} \) which becomes small as \(x\) increases. This indicates potential approaches zero because the effects of each charge decrease at large distances.
Key Concepts
Coulomb's Law: Understanding the Force Between ChargesElectric Field: The Force Field Around ChargesPoint Charges: The Basic Building Blocks of Electric Fields
Coulomb's Law: Understanding the Force Between Charges
Coulomb's Law is a cornerstone principle in the study of electromagnetism. This law describes the force between two point charges. According to Coulomb's Law, the electric force (\( F \)) between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, it’s expressed as:
Coulomb's Law helps us understand that:
- \( F = \frac{k |q_1 q_2|}{r^2} \)
Coulomb's Law helps us understand that:
- Like charges repel each other.
- Opposite charges attract each other.
Electric Field: The Force Field Around Charges
An electric field is essentially a region around a charged particle where other charges would experience a force. It is a vector field, meaning it has both magnitude and direction. We use the symbol \( E \) to denote electric field, and the formula for calculating it due to a point charge is:
The electric field at any point in space is defined as the force experienced by a positive test charge placed at that point divided by the magnitude of the charge. Importantly, the direction of \( E \) is radial relative to the charge:
- \( E = \frac{kq}{r^2} \)
The electric field at any point in space is defined as the force experienced by a positive test charge placed at that point divided by the magnitude of the charge. Importantly, the direction of \( E \) is radial relative to the charge:
- Outward for positive charges.
- Inward for negative charges.
Point Charges: The Basic Building Blocks of Electric Fields
Point charges are hypothetical charges that occupy a single point in space with no actual size or structure. They act as simplifications to model how charges interact, as seen in the exercise where a point charge represents each charge located in space. The simplicity of point charges allows us to easily calculate the electric field, potential, and forces without worrying about the shape or volume of the charge.
In the context of the exercise, point charges simplify the task of calculating the electric potential on the x-axis. Given their positions at \((0,0)\) and \((a,0)\), these charges produce distinct effects in terms of the field and potential, directly governed by their magnitudes and positions:
In the context of the exercise, point charges simplify the task of calculating the electric potential on the x-axis. Given their positions at \((0,0)\) and \((a,0)\), these charges produce distinct effects in terms of the field and potential, directly governed by their magnitudes and positions:
- A positive point charge produces a field pointing outwards, reducing potential.
- A negative point charge creates a field pointing inwards, increasing potential on nearby points relative to the positive charge.
Other exercises in this chapter
Problem 19
Two point charges \(q_{1}=+2.40 \mathrm{nC} \quad\) and \(\quad q_{2}=\) \(-6.50 \mathrm{nC}\) are 0.100 \(\mathrm{m}\) apart. Point \(A\) is midway between the
View solution Problem 20
A positive charge \(+q\) is located at the point \(x=0\) \(y=-a,\) and a negative charge \(-q\) is located at the point \(x=0,\) \(y=+a .(\) a) Derive an expres
View solution Problem 23
(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 potent
View solution Problem 24
At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 \(\mathrm{V}\) and \(12.0 \mathrm{V} / \mathrm
View solution