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Chapter 3: Problem 13

Three charges \(+Q\) are situated at three corners of a square of side \(a\). Find the potential at the fourth corner. What would be the potential energy of a charge \(+Q\) placed at the fourth corner? What would be the potential energy of the system of four charges?

Short Answer

Expert verified
Answer: The potential at the fourth corner is given by \(V_{total}=kQ\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{a\sqrt{2}}\right)\). The potential energy of a +Q charge at the fourth corner is given by \(U=Q\left(kQ\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{a\sqrt{2}}\right)\right)\). The potential energy of the system of four charges is given by \(U_{system}=kQ^2\left(3\frac{1}{a}+2\frac{1}{a\sqrt{2}}+\frac{1}{2a}\right)\).

Step by step solution

01

Calculate the distance between the charges

Each charge is at a corner of a square with side \(a\). The horizontal and vertical separation between each charge is \(a\). The diagonal separation between the charges is the hypotenuse of a right triangle with sides \(a\), which can be found using the Pythagorean theorem: \(d=\sqrt{a^2+a^2}=\sqrt{2a^2}=a\sqrt{2}\).
02

Calculate the potential at the fourth corner

The electric potential \(V\) at a point due to a point charge is given by the formula \(V=\frac{kQ}{r}\), where \(k\) is the electrostatic constant, \(Q\) is the point charge, and \(r\) is the distance between the charge and the point. In our case, we have three charges, so we will calculate the potential due to each charge and sum them up. The potential at the fourth corner \(V_{total}\) is given by: \(V_{total}=\frac{kQ}{a}+\frac{kQ}{a}+\frac{kQ}{a\sqrt{2}}\) Now, factor out \(kQ\) to simplify the expression: \(V_{total}=kQ\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{a\sqrt{2}}\right)\)
03

Calculate the potential energy of a +Q charge at the fourth corner

The potential energy \(U\) of a charge \(+Q\) placed at the fourth corner is given by the formula \(U=QV\). Using the potential we found in step 2: \(U=Q\left(kQ\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{a\sqrt{2}}\right)\right)\)
04

Calculate the potential energy of the system of four charges

To find the total potential energy of the system, we must calculate the potential energy for every pair of charges. There are six pairs: three pairs with separation \(a\), two pairs with separation \(a\sqrt{2}\), and one pair with separation \(2a\). The potential energy of the system is given by: \(U_{system} = 3\left(\frac{kQ^2}{a}\right) + 2\left(\frac{kQ^2}{a\sqrt{2}}\right) + \left(\frac{kQ^2}{2a}\right)\) Simplify the expression: \(U_{system}=kQ^2\left(3\frac{1}{a}+2\frac{1}{a\sqrt{2}}+\frac{1}{2a}\right)\) Now we have the potential at the fourth corner, the potential energy of a \(+Q\) charge placed at the fourth corner, and the potential energy of the system of four charges.

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Key Concepts

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

Electric Charges
Electric charges are the fundamental units of electricity and are the basis for understanding electric forces and fields. Charges come in two types: positive and negative. Like charges repel each other, while opposite charges attract. This interaction is quantified by Coulomb's law, which states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In the context of our exercise, we have three positive charges, denoted as \(+Q\), positioned at the corners of a square. The interactions between these charges determine the electric potential and potential energy at any point in the vicinity, including the fourth corner of the square. Understanding the behavior of these charges and the resulting electric field is critical in solving problems related to electric potential and potential energy.
Electric Potential Energy
Electric potential energy is the energy a charge possesses due to its position in an electric field. This is analogous to gravitational potential energy, where the energy depends on an object's height in a gravitational field. When charges are placed near other charges, they have the potential to do work as a result of the electrostatic forces acting upon them.

In our exercise, the potential energy of a charge \(+Q\) placed at the fourth corner can be derived from the overall electric potential created by the other three charges. Following through the solution steps, we determine this energy by multiplying the charge by the electric potential at that point. The concept of potential energy helps us understand how much work is needed to bring a charge into a particular position within an electric field or how much work can be obtained if the charge moves within the field.
Electrostatic Constant
The electrostatic constant, also referred to as Coulomb's constant \(k\), is a fundamental constant in electromagnetism. Its value in a vacuum is approximately \(8.9875 \times 10^9 \text{ N m}^2\text{/C}^2\). This constant appears in Coulomb's Law, which characterizes the electrostatic force between two charges, and in equations for calculating electric potential and electric fields.

In solving our exercise, the electrostatic constant is used to calculate the potential and potential energy. The value of \(k\) tells us the strength of the electrostatic force and the resulting potential for a given configuration of charges, and plays a crucial role in all calculations involving electric charges. Without this constant, it would be impossible to accurately predict the behavior of charges and the electric forces between them.

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Most popular questions from this chapter

An electron starts from the cathode of a vacuum tube with negligible velocity. What are its kinetic energy and velocity at the anode if the potential difference between electrodes is \(100 \mathrm{~V}\) ? How many electrons per unit volume at the anode are there if \(1 \mathrm{~mA}\) is collected by every square \(\mathrm{cm}\) of electrode? Explain how it is that the current in such an electron beam is constant when vin Eq. (1.12) is increasing across the tube.

Point charges \(+Q\) and \(-Q\) are separated by a distance \(2 l\), and \(O\) is the point midway between them. Find the E-field at \(P\), a distance \(r\) from \(O\), (a) when OP is along the line joining the charges and \(r>I\) and (b) when OP is perpendicular to the line joining the charges. Show that when \(r D\) l both fields are nearly proportional to \(Q l / r^{3}\) - (The charges form a dipole and \(\mathrm{P}\) is in the end-on and broadside positions, respectively -see Sec. 3.9.).

Find the relation between velocity and potential difference using classical mechanics for (a) electrons and (b) protons accelerated in vacuo. For what voltages do the velocities become onc-tenth of the velocity of light?

A large horizontal plane sheet with a uniform surface density of charge \(\sigma\) has a circular hole of radius \(b\) cut in it. If the charge on the remainder of the sheet stays uniformly distribated, find the E-field at a point a distance a vertically above the centre of the hole. (Long method: integration. Short method: superposition.)

A charge \(Q\) is spread uniformly round a thin wire bent into a circle O radius a. \(\mathrm{P}\) is a point on its axis at a distance \(r\) from the wire itself and \(\theta\) is the angle between \(r\) and the axis. Show that the E-field at P has a magnitude \(Q \cos \theta /\left(4 \pi e_{0} r^{2}\right)\). Show from that result that the work done in taking unit charge from \(O\) to infinity is \(Q /\left(4 \pi \epsilon_{0} a\right)\),
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