Three identical point charges q are placed at each of three corners of a square of side L. Find

step 1 So let's first draw the free body diagram to the problem. So we have three charges placed at thr

Three identical point charges q are placed at each of three...

Key Concepts

Coulomb's Law

Coulomb's Law describes the electrostatic force between two point charges. It states that the magnitude of the force is directly proportional to the product of the magnitudes of the two charges and inversely proportional to the square of the distance between them. This law is fundamental in calculating the forces between individual charges, taking into account whether the charge pair is attractive or repulsive based on their signs.

Superposition Principle

The superposition principle is essential for analyzing systems with multiple charges. It states that the net force acting on a charge is the vector sum of the individual forces exerted on it by all other charges. This principle allows one to compute the contributions from each charge independently and then combine them, a process that is crucial when dealing with configurations of several charges.

Symmetry in Electric Forces

Symmetry plays a significant role in simplifying the evaluation of net forces in charge distributions. When charges are arranged symmetrically, certain components of the forces can cancel out, reducing the complexity of the analysis. For instance, in symmetric configurations like a centered charge in a square, the equal and opposite force components along the same lines can lead to a net force of zero in specific directions.

Free-body Diagrams

Free-body diagrams are graphical representations used to visualize all the forces acting on a particular object. In the context of electric forces, drawing a free-body diagram helps in identifying the magnitude and direction of each force acting on a charge due to other charges. This method aids in the correct application of vector addition, ensuring that the net force is accurately determined.

Transcript

00:01 So let's first draw the free body diagram to the problem.

00:07 So we have three charges placed at three corners of a square.

00:20 And the length of the square, the length of the side of the square is l.

00:29 And in the first part, we have another charge, another point charge, negative three times q, placed at the center of.

00:40 The square.

00:42 So let's call this q prime and q prime is equal to negative three times q this is placed at the center of the square and now we have to show all the forces acting on discharge so since q dash is negative therefore all the forces acting on it should be attractive.

01:16 So the forces to capital q, sorry, the forces to do any of these charges will be towards that charge since the force is attractive.

01:27 So due to this force, this one, it should be f.

01:37 And since all of these charges have the same magnitude of the force due to them will also be safe.

01:42 And since all of these charges have the same magnitude, the magnitude of the force, due to them will also be safe.

01:48 So this will be the other corner will also be f and will be towards that corner and similarly for the third corner.

02:01 So these two forces as you can see that act due to the charges on the diagonal, so they are exactly aligned with the diagonal and are opposite to each other.

02:16 So this means that these two forces cancel each other out.

02:20 So the net force acting on q dash is just due to the charge in between the other two charges.

02:30 So the magnitude of that force will be k times q times q dash over square of this length.

02:41 And we know that the length of the diagonal of a square is root over of two times length of the length of the length.

02:55 Square length of each side.

02:58 So this distance is half of the length of the diagonal.

03:02 So root over of 2 over 2 times l and we need to take the square of that.

03:11 So this is equal to k times q times negative 3 q over l over root 2 whole square.

03:23 And let's substitute k.

03:26 So k is actually 1 over 4 by epsilon not.

03:29 And here we have negative 3 q squared and here we have l square over 2 so this gives the net force to have a magnitude 3 q square over 2 pi epsilon 0 so we have uh including the sign means uh we have considered the direction as well but here we have not yet choose a sorry, the sign actually specifies whether the force is attractive or repulsive.

04:13 And since we have already considered that while solving this problem, so let's just substitute the magnitude of the force without considering the sign...

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