Electric interaction

What Nour says

1. 1. Coulomb's law

   Today, the whole chapter: "Electric interaction", Year 3, Mathematics section.

   We will build this step by step, from one simple force to a full field picture.

   We start with Coulomb's law, then move to fields, point charges, two charges, and end with a uniform field.

   Let's start with Coulomb's law.

   This law describes the force between two electric charges.

   In other words, it tells us how strongly two charged objects push or pull each other.

   Picture two small charged balls, charge q1 and charge q2, separated by a distance r.

   Here is the second charge, placed a distance r away.

   The distance between their centers is called r, measured in meters, symbol m.

   Each charge creates an electric force on the other one, called the force électrique.

   That is, a push or a pull that one charge exerts on another, without touching it.

   Coulomb's law gives the strength of this force with a precise formula.

   Here F is the force, measured in newtons, symbol N.

   q1 and q2 are the two charges, each measured in coulombs, symbol C.

   r is the distance between the charges, measured in meters, symbol m.

   k is a constant number, called the Coulomb constant.

   This force acts along the straight line joining the two charges.

   Here is why the formula makes sense in a simple way.

   When the charges are bigger, the push or pull between them is stronger.

   When the charges are farther apart, the force becomes much weaker.

   That is why r is squared on the bottom of the formula.

   If charges have the same sign, they repel, pushing away from each other.

   If charges have opposite signs, they attract, pulling toward each other.

   Now let's solve a full example, like in the bac exam.

   Two point charges, q1 equals 2 microcoulombs and q2 equals minus 3 microcoulombs.

   They sit a distance r equals 0{,}3 meters apart in air.

   We want the magnitude of the electric force between them.

   step: {"n":1,"text":"Write Coulomb's law"}

   Step one, write Coulomb's law with the known values.

   step: {"n":2,"text":"Substitute the values"}

   Step two, substitute the charges in coulombs and the distance in meters.

   Multiply the two charges together, then take the absolute value.

   Divide by r squared, which is 0{,}09 square meters.

   step: {"n":3,"text":"Compute the result"}

   Multiply everything together, step by step, carefully.

   The force between the two charges is 0{,}6 newtons.

   Since the charges have opposite signs, this force is attractive.

   So remember, Coulomb's law links force, charges, and distance with one formula.

   The force grows with bigger charges and shrinks fast with more distance.

   Same signs push apart, opposite signs pull together.

   If the distance between two charges doubles, what happens to the force?

2. 2. The idea of electric field

   Now let's move to the idea of electric field.

   An electric field is a way to describe the effect of a charge on the space around it.

   That is, a charge changes the space near it, so another charge there feels a force.

   Picture one charge, called Q, sitting alone in space.

   This charge Q creates an electric field all around it, even with no second charge present.

   We picture this field with lines, called the field pattern and field lines (spectre et lignes de champ).

   That is, a picture made of lines showing the direction the field points at each spot.

   Each line points away from Q, since Q here is a positive charge.

   If we place a tiny test charge at any point, the field tells us the force on it.

   At point M, the field has a size and a direction, like an arrow.

   We write this field as E, measured in volts per meter, symbol V/m.

   Here is why we need this idea of field at all.

   Without field, we can only talk about force between two charges that are both present.

   With field, we describe the space around one charge, before any second charge arrives.

   This makes it easier to study many situations with the same idea.

   Let's work through an example close to the bac exam style.

   A point M sits near a positive charge Q, inside its electric field.

   A small test charge q0 is placed at point M, with q0 positive too.

   The field at M points away from Q, since Q is positive.

   step: {"n":1,"text":"Find the field direction"}

   Step one, since Q is positive, the field vector E at M points outward, away from Q.

   step: {"n":2,"text":"Find the force direction"}

   Step two, the force on q0 is F equals q0 times E.

   Since q0 is positive, the force F has the same direction as the field E.

   So the force on this test charge also points away from Q.

   This relation between force and field is a key tool for every case ahead.

   So remember, a charge creates a field around it, even alone in space.

   The field tells us the force that any other charge would feel there.

   Field lines show the field's direction, pointing away from positive charges.

   Does the field exist at point M before we place a test charge there?

3. 3. Electric field from a point charge

   Now let's study the électric field created by a point charge.

   That is, the exact field made by one single small charged object, at any distance from it.

   We already saw the field lines around a charge Q, pointing outward.

   Now we need a formula for the size of this field at any point.

   Take a point M, at a distance r from the charge Q.

   The field at M is given by this formula.

   Here E is the field, measured in volts per meter, symbol V/m.

   Q is the charge creating the field, measured in coulombs, symbol C.

   r is the distance from Q to the point M, measured in meters, symbol m.

   k is the same Coulomb constant we used before.

   The field vector at M points away from Q if Q is positive.

   It points toward Q if Q is negative.

   Here is why this formula looks so close to Coulomb's law.

   The force on a test charge q0 at M is F equals q0 times E.

   Combine that with Coulomb's law, and the field formula naturally appears.

   Farther from Q, the field gets weaker, following the same inverse square pattern.

   Let's solve a full example, in the bac exam style.

   A point charge Q equals 5 microcoulombs sits alone in space.

   We want the field E at a point M, 0{,}2 meters away from Q.

   step: {"n":1,"text":"Write the formula"}

   Step one, write the field formula with the known values.

   step: {"n":2,"text":"Substitute values"}

   Step two, put in Q in coulombs and r in meters.

   Square the distance first, giving 0{,}04 square meters.

   step: {"n":3,"text":"Compute"}

   Multiply the top first, then divide by 0{,}04.

   The field at M is 1{,}125 times 10 to the 6 volts per meter.

   Since Q is positive, this field points away from the charge.

   So remember, the field from a point charge follows an inverse square formula.

   It uses the same Coulomb constant k as the force law.

   Direction depends on the sign of the charge creating it.

   If the distance r is tripled, what happens to the field E?

4. 4. Electric field from two point charges

   Now let's look at the case of two point charges together.

   That is, when two charges both make their own field, and we combine them at one point.

   Picture two charges, Q1 and Q2, placed apart from each other.

   We want the total field at a point M, somewhere between or near them.

   Q1 alone creates a field E1 at M, and Q2 alone creates a field E2 at M.

   The true field at M is the sum of these two vectors, E1 plus E2.

   We write this as a vector sum, not just adding sizes directly.

   Here is why we must add them as vectors, with direction.

   Each field has its own direction, from E1 and E2 at different angles.

   We cannot just add their sizes, since direction changes the total result.

   Only when both fields point the same way can we add sizes directly.

   Let's solve a bac style example with two charges.

   Charge Q1 equals 4 microcoulombs and charge Q2 equals minus 4 microcoulombs.

   Point M sits on the line joining them, 0{,}1 meters from each charge.

   step: {"n":1,"text":"Find each field's size"}

   Step one, since both charges have equal size and equal distance, E1 equals E2.

   Compute this value, giving 3{,}6 times 10 to the 6 volts per meter.

   step: {"n":2,"text":"Find the directions"}

   Step two, Q1 is positive, so E1 points away from Q1, toward M's right side.

   Q2 is negative, so E2 points toward Q2, in the same direction as E1 here.

   step: {"n":3,"text":"Add the fields"}

   Step three, since E1 and E2 point the same way here, we just add their sizes.

   The total field at M is 7{,}2 times 10 to the 6 volts per meter.

   So remember, with two charges, each creates its own field at the point M.

   We combine both fields as vectors, respecting their direction, not just their size.

   Only aligned fields let us simply add or subtract their sizes directly.

   If E1 and E2 point in opposite directions, how do we combine them?

5. 5. Uniform electric field

   Now let's finish with the uniform electric field.

   That is, a field that has the same size and the same direction at every point.

   This happens between two flat parallel plates, charged with opposite signs.

   Between the plates, the field lines are straight and all parallel to each other.

   Every line points from the positive plate toward the negative plate.

   We give the field's size with a simple formula using the voltage U.

   Here E is the field, measured in volts per meter, symbol V/m.

   U is the voltage between the plates, measured in volts, symbol V.

   d is the distance between the plates, measured in meters, symbol m.

   Here is why this field stays the same everywhere between the plates.

   Far from the edges, the plates look infinite, so the field has no favored point.

   This makes the force on any charge placed there always the same.

   That force is still F equals q times E, just like before.

   Let's solve a full example, in the bac exam style.

   Two parallel plates are 0{,}05 meters apart, symbol d.

   The voltage between them is 2000 volts, symbol U.

   step: {"n":1,"text":"Write the formula"}

   Step one, write the uniform field formula with U and d.

   step: {"n":2,"text":"Substitute values"}

   Step two, divide the voltage in volts by the distance in meters.

   step: {"n":3,"text":"Compute the result"}

   The field between the plates is 4 times 10 to the 4 volts per meter.

   A positive test charge placed here feels a force toward the negative plate.

   So remember, a uniform field has constant size and constant direction everywhere.

   It appears between two parallel charged plates, given by U divided by d.

   This closes our full tour of electric interaction, from force to field.

   In a uniform field, does the force on a charge change from point to point?

   That completes our full tour of electric interaction, from one simple force to a uniform field.

   You now have the formulas, the units, and the steps to solve any bac-style problem on this chapter.

   Keep practicing with real exercises, and these ideas will become second nature.