Three point charges are fixed in place in a right triangle, as shown in the figure. What is the

Three point charges are fixed in place in a right triangle,...

Key Concepts

Vector Addition of Forces

When forces are vectors, they have both magnitude and direction. Combining forces requires vector addition, which may involve decomposing each force into its components along chosen coordinate axes and then summing these components. This method is crucial for determining the net force acting on a charge in a multi-charge system.

Geometric Decomposition

Geometric decomposition involves breaking down vectors into their horizontal and vertical components using trigonometric functions. In problems where charges are arranged in shapes like a right triangle, this technique simplifies the process of adding forces by aligning them with the coordinate axes and accurately computing the resultant force.

Coulomb's Law

Coulomb's Law describes the force between two point charges. It states that the magnitude of the electric force is proportional to the product of the charge magnitudes and inversely proportional to the square of the distance between them. This law is fundamental in calculating the interaction forces between charges in an electrostatic system.

Superposition Principle

The superposition principle asserts that in a system with multiple charges, the total electric force on a given charge is the vector sum of the individual forces exerted by each of the other charges. This principle allows the analysis of complex charge configurations by treating the effect of each charge independently.

Transcript

00:02 Hi, the three charged particles are kept at the vertices of an equilateral triangle, sorry, right angled triangle.

00:11 This is the perpendicular of this right angle triangle.

00:16 This is its base and then hypotenuse.

00:21 You can consider this base to be along x -axis, this perpendicular to be along y -axis.

00:28 Here this is x -axis this is y -axis the charge particle here this is a positive charge this is also a positive charge but the charge kept at this vertex here this is a negative charge suppose the points are a the vertices are a b and c here this charge is plus 0 .80 micawatt here this is plus 1 .0 microculum and here this is minus 0 .60 microculum.

01:09 The height of this triangle the perpendicular is 8 .0 centimeter and hypotenuse is 10 .0 centimeter.

01:21 So as ab is 8 .0 centimeter and b .c means hypotenuse is 10 .0 centimeter and b c means hypotenuse is 6, that is 10 .0 centimeters.

01:35 So this base using pythagoras theorem that will be given by square root of the square of bc minus the square of ab.

01:47 So here this is the square of 10 which is 100, the square of 8 which is 64.

01:53 So it will come out to be 36 centimeter means finally ac is 6 .0 centimeter.

02:01 Now we have to find the net force along with the direction acting at this charge which is kept at point c.

02:10 So there are two forces acting at this point c.

02:15 One is repulsion, the repulsive force exerted by 0 .80 microculum at this plus 1 .0 micropulum.

02:24 As this is repulsive so we can show it here in this direction and that is f at c due to b and another force attractive force, force of attraction exerted at 1 .0 microculum by this 0 .60 microculum.

02:42 So this is f at c again but due to a.

02:47 This will be the angle between them which we can assume to be theta and that theta will be 180 degree minus this alpha.

02:56 So, as we will have to find the vector sum of these two forces also for which we will need this theta.

03:04 So we find this theta in the beginning for which we will use here cos alpha we are using in this right angle triangle, using the trigonometric relation cos alpha which is base means 6 ac which we have found in it here.

03:24 So 6 .0 centimeter divided by hypotenous 10 .0 centimeter.

03:30 So this cause alpha here will come out to be 0 .6.

03:34 And that will be used later on.

03:37 Now, we will find the magnitude, first of all, of these two forces.

03:42 So fca, using coulin's law, this fca will be given by k into charge captate a into charge kept at c divided by the distance ac having a whole square now plugging in all known values for k this is 9 into 10 x per 9 into charge 0 .6 microculum or 0 .60 into 10 dash per minus 6 coulum multiplied by 1 microculum or 1 .0 into 10 dash per minus 6 column divided by a distance square means this is 6 cm or 6 .0 into 10 t .m.

04:21 Minus 2 meter to the whole square.

04:23 So finally this force comes out to be 1 .5 newton...

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