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A comprehensive guide to understanding and applying coulomb's law in electrostatics. It presents multiple examples of calculating the force between point charges in various configurations, including those on the x-axis and in two dimensions. The document also explores the concept of superposition and demonstrates how to determine the net force on a charge due to multiple other charges. It includes detailed explanations, step-by-step calculations, and diagrams to illustrate the principles involved. This resource is valuable for students studying electrostatics and related concepts in physics.
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Ejemplo 1
3 cargas puntuales Q1= Q2 = 2ฮผC Y Q3 = 4ฮผC estan colocadas como se muestra.
Encuentre la fuerza resultante en Q3.
13
1
2
2
13
9
โ 6
โ 6
2
13
13
23
13
๐ฅ
13
13
๐ฆ
13
๐ฅ
13
๐ฅ
23
๐ฅ
13
๐ฅ
13
๐ฆ
13
๐ฆ
23
๐ฆ
๐ฆ
13
23
Ejemplo 1 Metodo vectorial
1
2
3
Principio de superposicio n
๐ 3
3
1
3
1
3
1
3
3
2
3
2
3
2
3
๐ 3
3
1
3
1
3
1
3
2
3
2
3
2
3
๐ 3
13
23
3
1
3
2
3
1
3
2
2
2
๐ 3
9
๐ร๐
2
๐ถ
2
โ 6
( 2 ร 10
โ 6
๐ถ)
(
) ๐
( 0. 5 ๐)
3
( 2 ร 10
โ 6
๐ถ)
(
) ๐
( 0. 5 ๐)
3
๐ 3
๐ 3
๐ 3
b)
๐ 0
0
1
2
2
2
๐ 0
9
๐ร๐
2
๐ถ
2
โ 9
25 ๐ถ ๐ฬ
๐ฅ
2
10 ๐ถ ๐ฬ
( ๐ฅโ 2
)
2
โ 9
๐ 0
2
2
โ 9
๐ 0
2
2
โ 6
๐ 0
2
2
โ 6
๐ 0
2
2
โ 6
2
2
2
2
2
2
2
๐ 0
๐ 0
2
2
๐ 0
2
2
๐ 0
3
3
๐ 0
3
3
๐ 0
3
๐ 0
๐ 0
๐ 0
๐ 0
Ley de Coulomb
27.- Tres cargas puntuales estan en el eje x; ๐ 1
= โ 6 , 0 ๐๐ถ esta en x = - 3,0 m,
2
= 4 , 0 ๐๐ถ esta en el origen y ๐
3
= โ 6 , 0 ๐๐ถ esta en x = - 3,0 m. Hallar la fuerza
ejercida sobre ๐ 1
1
2
3
๐ 1
๐ 1
1
2
21
21
2
3
31
31
2
21
31
๐ 1
9
2
2
โ 9
โ 6
2
โ 6
2
๐ 1
โ 3
๐ 1
โ 3
๐ 1
โ 2
29.- Dos cargas puntuales de - 2 y 4 ฮผC, respectivamente, estan separadas una distancia
Lโฆ ยฟDonde se deberรญ a poner una tercera carga para que la fuerza electrica sobre ella
fuera nula?
๐ 0
0
2
21
21
2
3
31
31
2
10
20
10
20
๐ 0
0
2
2
โ 6
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
2
35.- Cinco cargas iguales Q estan igualmente espaciadas en un semicรญrculo de radio R
como indica la figura. Determinar la fuerza (en funcion de K, Q y R) que se ejerce sobre
una carga q localizada equidistante de las otras cargas en el centro del semicรญ rculo.
๐
๐
1
1
2
2
3
3
4
4
5
5
1
2
3
4
5
2
๐
๐
5
๐ = 1
1
2
3
4
5
๐
5
๐ = 1
๐
2
Ejercicio
๐
๐
๐
๐
3
5
๐ = 1
1
2
3
4
5
1
๐
1
๐
3
๐
5
๐ = 1
๐
3
๐
5
๐ = 1
๐
3
๐
3