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Chapter 23: Problem 18

Two charges of equal magnitude \(Q\) are held a distance \(d\) apart. Consider only points on the line passing through both charges. (a) If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity is zero (is the electric field zero at these points? (ii) the electric field is zero (is the potential zero at these points?). (b) Repeat part (a) for two charges having opposite signs.

Short Answer

Expert verified
Same sign: No zero potential points, zero field at center (\(x = 0\)). Opposite sign: Zero potential outside (\(x = \pm d\)), no zero field between charges.

Step by step solution

01

Understanding Potential with Same Sign Charges

Let's consider two charges, both with a magnitude of \(Q\) and located at positions \(x = -d/2\) and \(x = d/2\). We need to find points \(x\) where the potential \(V(x)\) is zero. The potential from a single charge at a distance \(r\) is \(V = \frac{kQ}{r}\) where \(k\) is Coulomb's constant. For two charges, the total potential is the sum of their individual potentials: \[V(x) = \frac{kQ}{|x + d/2|} + \frac{kQ}{|x - d/2|}.\] Set this equal to zero to find points where the potential is zero.
02

Solving for Zero Potential with Same Sign Charges

The equation \(\frac{kQ}{|x + d/2|} + \frac{kQ}{|x - d/2|} = 0\) simplifies to \(\frac{1}{|x + d/2|} + \frac{1}{|x - d/2|} = 0\). This is impossible for real values of \(x\) since both terms are positive. Thus, there are no points where the potential is zero.
03

Understanding Electric Field with Same Sign Charges

The electric field \(E(x)\) is given by the derivative of the potential. It's zero where the forces from both charges cancel: \[ E(x) = -\frac{kQ}{(x + d/2)^2} + \frac{kQ}{(x - d/2)^2} = 0. \] This requires that \(x\) is halfway between the charges where the forces are equal.
04

Solving for Zero Electric Field with Same Sign Charges

The equation \(-\frac{kQ}{(x + d/2)^2} + \frac{kQ}{(x - d/2)^2} = 0\) simplifies to \(\frac{1}{(x + d/2)^2} = \frac{1}{(x - d/2)^2}\), which gives \(x = 0\). At \(x = 0\), the field is zero, but the potential is not zero.
05

Potential for Opposite Sign Charges

For opposite sign charges, the potential zero points are found similarly: \[ V(x) = \frac{kQ}{|x + d/2|} - \frac{kQ}{|x - d/2|} = 0. \] From symmetry, this simplifies to finding where \(x\) is such that \(|x + d/2| = |x - d/2|\).
06

Solving for Zero Potential with Opposite Sign Charges

Solving \(\frac{1}{|x + d/2|} = \frac{1}{|x - d/2|}\), the potential is zero at \(x = \pm d\), found outside the region between the charges. Hence, potential is zero outside but never inside.
07

Electric Field for Opposite Sign Charges

For the electric field: \[ E(x) = -\frac{kQ}{(x + d/2)^2} - \frac{kQ}{(x - d/2)^2}, \] we want this equation to equal zero. This can be true only very far from the charges, so effectively not between the charges.
08

Conclusion for Opposite Sign Charges

For opposite charges, the electric field is never zero between them because they attract each other. The potential is zero at points outside the charges, specifically \(x = \pm d\).

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle in electrostatics. It describes the force between two point charges. The law states that the force,
  • is directly proportional to the product of the magnitudes of the charges,
  • is inversely proportional to the square of the distance between the charges,
  • acts along the line joining the two charges.
This can be expressed with the equation: \[ F = \frac{k |Q_1 Q_2|}{r^2} \]Here, \(k\) is Coulomb's constant, \(Q_1\) and \(Q_2\) are the magnitudes of the charges, and \(r\) is the distance between the charges. Coulomb's Law is crucial for calculating the electric forces and serves as a foundation for understanding electric fields and potential.
Same Sign Charges
Charges with the same sign (both positive or both negative) repel each other. This principle is vital when calculating electric fields and potentials.
In this scenario, it's key to understand:
  • The electric potential (energy per unit charge due to a field) from each charge sums up.
  • The electric potential from two same sign charges can never be zero at a point, since potentials from same sign charges add up and can't cancel to zero.
  • However, the electric field can be zero at a point where force vectors from both charges cancel each other out completely.
For two charges placed at a distance apart, the electric field zero point is exactly midway between them, where the forces balance.
Opposite Sign Charges
Charges with opposite signs attract each other. Their interaction affects both the electric potential and electric field notably.
With opposite sign charges:
  • The potentials from these charges can actually cancel out at certain points, resulting in zero potential locations.
  • These zero potential points are not between the charges, but rather outside the space between them.
  • Conversely, the electric field never reaches zero between opposite charges due to the attraction causing a net force.
This concept is integral when studying fields due to dipoles or understanding molecular interactions.
Electrostatics Problems
Handling electrostatics problems typically involves understanding the behavior and effects of electric charges.
Some important aspects include:
  • Using Coulomb's Law to determine forces between charges.
  • Calculating the electric field at various points due to a given configuration of charges.
  • Understanding the potential energy and resulting potential differences which can determine the work done by an electric field.
Solving these problems often requires conceptual understanding as well as mathematical manipulation. Being able to visualize the setup and resulting fields or potentials helps in developing an intuitive approach to problem-solving.

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Most popular questions from this chapter

A small metal sphere, carrying a net charge of \(q_{1}=\) \(-2.80 \mu \mathrm{C},\) is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of \(q_{2}=-7.80 \mu \mathrm{C} \quad\) and \(\operatorname{mass}\) \(1.50 \mathrm{g},\) is projected toward \(q_{1}\). When the two spheres are 0.800 \(\mathrm{m}\) apart, \(q_{2}\) is moving toward \(q_{1}\) with speed 22.0 \(\mathrm{m} / \mathrm{s}\) (Fig. E23.5). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. (a) What is the speed of \(q_{2}\) when the spheres are 0.400 \(\mathrm{m}\) apart? (b) How close does \(q_{2}\) get to \(q_{1}\) ?

Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 \(\mathrm{cm} .\) (a) If the surface charge density for each plate has magnitude 47.0 \(\mathrm{nC} / \mathrm{m}^{2}\) , what is the magnitude of \(\vec{\boldsymbol{E}}\) in the region between the plates? (b) What is the potential difference between the two plates? (c) If the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field and to the potential difference?

An alpha particle with kinetic energy 11.0 MeV makes a head-on collision 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 that it may be treated as a point charge. The atomic number of lead is \(82 .\) The alpha particle is a helium nucleus, with atomic number \(2.)\)

Four electrons are located at the corners of a square 10.0 \(\mathrm{nm}\) on a side, with an alpha particle at its midpoint. How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

A ring of diameter 8.00 \(\mathrm{cm}\) is fixed in place and carries a charge of \(+5.00 \mu \mathrm{C}\) uniformly spread over its circumference. (a) How much work does it take to move a tiny \(+3.00-\mu C\) charged ball of mass 1.50 \(\mathrm{g}\) from very far away to the center of the ring? (b) Is it necessary to take a path along the axis of the ring? Why? (c) If the ball is slightly displaced from the center of the ring, what will it do and what is the maximum speed it will reach?
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