Coulomb's Law: Exercises

Coulomb's law is used to calculate the magnitude of the electrical force between two charges.

This law says that the force intensity is equal to the product of a constant, called constant electrostatics, by the modulus of the value of the charges, divided by the square of the distance between the charges, i.e:

Take advantage of the resolution of the questions below to clear your doubts regarding this electrostatic content.

Resolved issues

1) Fuvest - 2019

Three small spheres charged with a positive charge ܳ occupy the vertices of a triangle, as shown in the figure. In the inner part of the triangle is affixed another small sphere, with a negative charge q. The distances of this charge to the other three can be obtained from the figure.

Where Q = 2 x 10^-4 C, q = - 2 x 10^-5 C and ݀d = 6 m, the net electric force on the charge q

(The constant k₀ Coulomb's law is 9 x 10⁹ No. m² /Dz)

a) is null.
b) has y-axis direction, downward direction and 1.8 N modulus.
c) has y-axis direction, upward direction and 1.0 N modulus.

d) has y-axis direction, downward direction and 1.0 N modulus.
e) has y-axis direction, upward direction and 0.3 N module.

See Answer

To calculate the net force on the load q it is necessary to identify all the forces acting on this load. In the image below we represent these forces:

Charges q and Q1 are located at the vertex of the right triangle shown in the figure, which has legs measuring 6 m.

Thus, the distance between these charges can be found through the Pythagorean theorem. So we have:

Now that we know the distances between the charges q and Q₁, we can calculate the strength of the force F₁ among them applying Coulomb's law:

The strength of the F force₂ between q and q charges₂ will also be equal to , because the distance and the value of the charges are the same.

To calculate the net force F₁₂ we use the parallelogram rule, as shown below:

To calculate the force value between q and Q loads₃ we again apply Coulomb's law, where the distance between them is equal to 6 m. Thus:

Finally, we will calculate the net force on the charge q. Note that the F forces₁₂ and F₃ have the same direction and opposite direction, so the resulting force will be equal to the subtraction of these forces:

How F₃ has a module greater than F₁₂, the result will point up in the y-axis direction.

Alternative: e) has y-axis direction, upward direction and 0.3 N modulus.

To learn more, see Coulomb's Law and electric power.

2) UFRGS - 2017

Six electrical charges equal to Q are arranged, forming a regular hexagon with edge R, as shown in the figure below.

Based on this arrangement, with k being the electrostatic constant, consider the following statements.

I - The resulting electric field in the center of the hexagon has a modulus equal to
II - The work required to bring a charge q, from infinity to the center of the hexagon, is equal to
III - The resultant force on a test load q, placed in the center of the hexagon, is null.

Which ones are correct?

a) Only I.
b) Only II.
c) Only I and III.
d) Only II and III.
e) I, II and III.

See Answer

I - The electric field vector in the center of the hexagon is null, as the vectors of each charge have the same modulus, they cancel each other out, as shown in the figure below:

So the first statement is false.

II - To calculate the work we use the following expression T = q. ΔU, where ΔU is equal to the potential at the center of the hexagon minus the potential at infinity.

Let's define the potential at infinity as null and the value of the potential at the center of the hexagon will be given by the sum of the potential relative to each charge, since the potential is a scalar quantity.

Since there are 6 charges, then the potential at the center of the hexagon will be equal to: . In this way, the work will be given by: , therefore, the statement is true.

III - To calculate the net force at the center of the hexagon, we do a vector sum. The resultant force value at the center of the hex will be zero. So the alternative is also true.

Alternative: d) Only II and III.

To learn more, see also Electric field and Electric Field Exercises.

3) PUC/RJ - 2018

Two electrical charges +Q and +4Q are fixed on the x-axis, respectively at positions x = 0.0 m and x = 1.0 m. A third charge is placed between the two, on the x-axis, such that it is in electrostatic equilibrium. What is the position of the third charge, in m?

a) 0.25
b) 0.33
c) 0.40
d) 0.50
e) 0.66

See Answer

When placing a third load between the two fixed loads, regardless of its sign, we will have two forces of the same direction and opposite directions acting on this load, as shown in the figure below:

In the figure, we assume that charge Q3 is negative and since the charge is in electrostatic equilibrium, then the net force is equal to zero, like this:

Alternative: b) 0.33

To learn more, see electrostatics and Electrostatics: Exercises.

4) PUC/RJ - 2018

A load that₀ is placed in a fixed position. When placing a load q₁ =2q₀ at a distance d from q₀, what₁ suffers a repulsive force of modulus F. Replacing q₁ for a load that₂ in the same position, which₂ suffers an attractive force of 2F modulus. If the loads q₁ and what₂ are placed at a 2d distance from each other, the force between them is

a) repulsive, of module F
b) repulsive, with a 2F module
c) attractive, with module F
d) attractive, with 2F module
e) attractive, 4F module

See Answer

As the force between the charges q_O and what₁ is repulsion and between charges q_O and what₂ is of attraction, we conclude that the loads q₁ and what₂ have opposite signs. In this way, the force between these two charges will be of attraction.

To find the magnitude of this force, we will start by applying Coulomb's law in the first situation, that is:

Being the load q₁ = 2 q₀the previous expression will be:

When replacing q₁ why₂ the force will be equal to:

Let's isolate the charge that₂ on a two sides of the equality and replace the value of F, so we have:

To find the net force between the charges q₁ and what₂, let's apply Coulomb's law again:

Replacing q₁ for 2q₀, what₂ by 4q₀ and of₁₂ by 2d, the previous expression will be:

Observing this expression, we notice that the module of F₁₂ = F.

Alternative: c) attractive, with module F

5) PUC/SP - 2019

A spherical particle electrified with a charge of modulus equal to q, of mass m, when placed on a flat, horizontal, perfectly smooth surface with its center a a distance d from the center of another electrified particle, fixed and also with a charge of modulus equal to q, is attracted by the action of the electric force, acquiring an acceleration α. It is known that the electrostatic constant of the medium is K and the magnitude of the acceleration of gravity is g.

Determine the new distance d’, between the centers of the particles, on this same surface, however, with it now inclined at an angle θ, in relation to the horizontal plane, so that the load system remains in balance static:

See Answer

For the load to remain in equilibrium on the inclined plane, the component of the force weight must be in the direction tangent to the surface (P_t ) is balanced by electrical force.

In the figure below we represent all the forces acting on the load:

The P component_t of the weight force is given by the expression:

P_t = P. if not

The sine of an angle is equal to the division of the measure of the opposite leg by the measure of the hypotenuse, in the image below we identify these measures:

From the figure, we conclude that sen θ will be given by:

Substituting this value in the weight component expression, we are left with:

As this force is being balanced by the electrical force, we have the following equality:

Simplifying the expression and isolating the d', we have:

Alternative:

6) UERJ - 2018

The diagram below represents the metallic spheres A and B, both with masses of 10^-3 kg and electrical load of module equal to 10^-6 Ç. The spheres are attached by insulating wires to supports, and the distance between them is 1 m.

Assume that the wire holding sphere A has been cut and that the net force on that sphere corresponds only to the electrical interaction force. Calculate the acceleration, in m/s², acquired by ball A immediately after cutting the wire.

See Answer

To calculate the value of the acceleration of the sphere after cutting the wire, we can use Newton's 2nd law, ie:

F_R = m. The

Applying Coulomb's law and equating the electric force to the resulting force, we have:

Replacing the values ​​indicated in the problem:

7) Unicamp - 2014

The attraction and repulsion between charged particles has numerous industrial applications, such as electrostatic painting. The figures below show the same set of charged particles, at the vertices of a square side a, that exert electrostatic forces on charge A at the center of this square. In the situation presented, the vector that best represents the net force acting on load A is shown in figure

See Answer

The force between charges of the same sign is attraction and between charges of opposite signs is repulsion. In the image below we represent these forces:

Alternative: d)