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Four identical charges \(Q\) are placed at the corners of a square of side \(L\). (a) In a free-body diagram, show all of the forces that act on one of the charges. (b) Find the magnitude and direction of the total force exerted on one charge by the other three charges.

Short Answer

Expert verified
The magnitude of the total force exerted on one charge by the three other charges is \(F_{total} = K \cdot \frac{{Q^2}}{{L^2}} \cdot (\sqrt{2} + 1/2)\) directed along the line joining the charge under consideration and the charge diagonally opposite to it.

Step by step solution

01

Draw a Free Body Diagram

In a square layout, there are four charges at the corners. Select one charge and draw the forces that act on it due to the other three charges. The force \(F\) between any two charges is given by Coulomb's law, \(F = K \cdot \frac{{Q^2}}{{r^2}} \), where \(r\) is the distance between the charges and \(K\) is Coulomb's constant. Remember, electric forces obey the inverse square law. For each charge, there will be a force directed towards the other charge. Note that angles will be important for the next steps, so ensure these are drawn in the diagram.
02

Calculate the Force Due to Each Charge

Calculate the force due to charges that are next to the selected charge. These charges are at a distance \(L\) from the selected charge. The force due to each of these charges is \(F_{adjacent} = K \cdot \frac{{Q^2}}{{L^2}}\). Calculate the force due to charge that is diagonally across from the selected charge. It is at a distance \(\sqrt{2 L}\) from the selected charge due to Pythagorean theorem. So, the force due to this charge is \(F_{diagonal} = K \cdot \frac{{Q^2}}{{2L^2}}\).
03

Find the Resultant Force

The vector sum of these three forces gives the total force on the charge. The forces from the adjacent charges have a resultant \(F_{adjacent}= 2 F_{adjacent} cos(45°)\) in the direction perpendicular to the diagonal. Add this to the force from the diagonal charge to get the total force, \(F_{total} = F_{adjacent} + F_{diagonal} = K \cdot \frac{{Q^2}}{{L^2}} \cdot (\sqrt{2} + 1/2)\).

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Key Concepts

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

Electric Force
Understanding the electric force is fundamental in the realm of electrostatics, the study of stationary electric charges. Electric force is a vector quantity, which means it has both magnitude and direction, and it arises due to the presence of electric charges. It can be attractive or repulsive depending on whether the charges in question are opposite or like, respectively.

When considering electric forces between point charges, Coulomb's Law provides the equation necessary to calculate this force: \[ F = K \cdot \frac{{Q^2}}{{r^2}} \], where \(F\) is the magnitude of the force, \(K\) is Coulomb's constant (approximately \(8.99 \times 10^9\) Nm^2/C^2), \(Q\) is the magnitude of each charge, and \(r\) is the distance between the charges. This equation highlights that the force is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them, an example of the inverse square law.

In the given exercise, this law is used to calculate the forces acting on one charge due to the other three placed at the corners of a square. Each pairwise interaction is governed by Coulomb's Law, leading to forces that are central—directly along the line connecting the two charges.
Free Body Diagram
A free body diagram is a powerful tool used in physics to visualize the forces acting on an object and how they influence its motion. In essence, it's a simple drawing that represents the object of interest as a dot or a small box and uses arrows to depict the various forces exerted on it. Each arrow's length symbolizes the magnitude of force, while the direction indicates the direction in which the force is applied.

For electric forces, the free body diagram enables us to see how forces due to electric charges interact. To create an accurate free body diagram of a charge in a system of multiple charges, like the square arrangement described in the exercise, one must draw all the electric forces acting on the selected charge due to the presence of other charges. It's essential to represent both the magnitude and the direction correctly to understand how these forces balance or combine to affect the charge's motion.

Completing such a diagram is not merely an exercise in illustration but the basis for further calculation and understanding. By breaking down complex force interactions into manageable components, the diagram makes it easier to apply principles like Coulomb's Law for each individual force involved.
Vector Sum of Forces
To analyze the combined effect of multiple forces acting on a single point, we must consider their vector sum or resultant force. This concept is especially relevant when dealing with electric forces because these forces can act in different directions and their combined effect determines the motion of the charge.

The vector sum is a method of adding multiple vectors, taking into account both their magnitudes and directions. When forces are not aligned—which is often the case with multiple charges, like in the exercise provided—the resultant force requires us to use vector addition principles, involving both the vertical and horizontal components of each force.

For the example of charges at the corners of a square, the forces from adjacent charges can be combined by finding their components along perpendicular axes and summing those to find the total force in each direction. The diagonal force is already along one axis, so it simplifies the calculation. Upon summing, the charge's movement will follow the direction of the resultant force, illustrating the collective influence of all individual forces at play.

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Most popular questions from this chapter

The ammonia molecule \(\left(\mathrm{NH}_{3}\right)\) has a dipole moment of \(5.0 \times 10^{-30} \mathrm{C} \cdot \mathrm{m} .\) Ammonia molecules in the gas phase are placed in a uniform electric field \(\vec{E}\) with magnitude \(1.6 \times 10^{6} \mathrm{~N} / \mathrm{C}\). (a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to \(\vec{E}\) from parallel to perpendicular? (b) At what absolute temperature \(T\) is the average translational kinetic energy \(\frac{3}{2} k T\) of a molecule equal to the change in potential energy calculated in part (a)? (Note: Above this temperature, thermal agitation prevents the dipoles from aligning with the electric field.)

(a) An electron is moving east in a uniform electric field of \(1.50 \mathrm{~N} / \mathrm{C}\) directed to the west. At point \(A,\) the velocity of the electron is \(4.50 \times 10^{5} \mathrm{~m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{~m}\) east of point \(A ?\) (b) A proton is moving in the uniform electric field of part (a). At point \(A,\) the velocity of the proton is \(1.90 \times 10^{4} \mathrm{~m} / \mathrm{s},\) east. What is the speed of the proton at point \(B ?\)

A small sphere with positive charge \(q\) and mass \(m\) is released from rest in a uniform electric field \(\vec{E}\) that is directed vertically upward. The magnitude of the field is large enough for the sphere to travel upward when it is released. How long does it take the sphere to travel upward a distance \(d\) after being released from rest? Give your answer in terms of \(q, m, d, E,\) and the acceleration due to gravity, \(g\).

Consider an infinite flat sheet with positive charge density \(\sigma\) in which a circular hole of radius \(R\) has been cut out. The sheet lies in the \(x y\) -plane with the origin at the center of the hole. The sheet is parallel to the ground, so that the positive \(z\) -axis describes the "upward" direction. If a particle of mass \(m\) and negative charge \(-q\) sits at rest at the center of the hole and is released, the particle, constrained to the \(z\) -axis, begins to fall. As it drops farther beneath the sheet, the upward electric force increases. For a sufficiently low value of \(m,\) the upward electrical attraction eventually exceeds the particle's weight and the particle will slow, come to a stop, and then rise back to its original position. This sequence of events will repeat indefinitely. (a) What is the electric field at a depth \(\Delta\) beneath the origin along the negative \(z\) -axis? (b) What is the maximum mass \(m_{\max }\) that would prevent the particle from falling indefinitely? (c) If \(m

A proton is placed in a uniform electric field of \(2.75 \times 10^{3} \mathrm{~N} / \mathrm{C} .\) Calculate (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after \(1.00 \mu \mathrm{s}\) in the field, assuming it starts from rest.

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