🎯 Key Takeaways for quick navigation: The video presents an electrostatics problem involving a triangular arrangement of charges. The goal is to determine the potential at the vertex (point C) and the potential at a specific point (point D), as well as the work done by the electric field to move a charge from point C to point D. The points A, B, and C represent the corners of an equilateral triangle with a side length of 100 millimeters. Charge q1 = +8 μC is located at point B, 100 millimeters to the left of charge q2 = -8 μC. Point D is located 20 millimeters to the left of charge q2. The potential at point C is zero volts (V = 0 V). This is because the net potential at a point due to multiple charges is the sum of the individual potentials due to each charge, and the contributions from q1 and q2 cancel each other out at point C. The potential at point D is -2.7 × 10^-6 volts (V = -2.7 × 10^-6 V). This is calculated using the equation V = kq/r, where V is the potential, k is Coulomb's constant, q is the charge, and r is the distance between the charge and the point. The work done by the electric field to move a charge of 2 μC from point C to point D is 5.4 joules (W = 5.4 J). This is calculated using the equation W = qΔV, where W is the work, q is the charge, and ΔV is the potential difference. The potential at point C is zero volts (V = 0 V). This is because the net potential at a point due to multiple charges is the sum of the individual potentials due to each charge, and the contributions from q1 and q2 cancel each other out at point C. Electric potential is a scalar quantity that represents the potential energy per unit charge at a given point in an electric field. It is measured in volts (V). The potential difference between two points is the change in potential energy per unit charge as a charge moves between those points. The electric field is a vector quantity that describes the force exerted on a unit charge at a given point. The resultant potential at point C is zero volts (V = 0 V). This is because the sum of the individual potentials due to the charges at points A and B is zero. The potential due to charge q1 at point C is positive, while the potential due to charge q2 at point C is negative. The magnitudes of these two potentials are equal, resulting in a net potential of zero at point C. The goal is to determine the potential at point D, which is located 20 millimeters to the left of charge q2. This involves identifying the distances between the charges and point D, applying the formula for electric potential, and calculating the resultant potential. The distance between point A and point D is assumed to be 80 millimeters. This is based on the fact that the triangle is equilateral and the total distance between point A and point C (which is also the distance between point B and point C) is 100 millimeters. The distance between point B and point D is 20 millimeters. The formula for electric potential (V) at a point due to a charge (q) is: V = k * q / r, where k is Coulomb's constant (9 × 10^9 N·m^2/C^2) and r is the distance between the charge and the point. Using this formula, calculate the potential (V_A) at point D due to charge q1 (8 μC) and the potential (V_B) at point D due to charge q2 (-8 μC). The resultant potential (V_D) at point D is the sum of the potentials (V_A and V_B) due to the individual charges: V_D = V_A + V_B. Substitute the calculated values for V_A and V_B into the equation to determine the resultant potential at point D. The goal is to determine the work done by the electric field to move a charge of 2 μC from point C to point D. This involves applying the formula for work done, which is W = qΔV, where W is the work, q is the charge, and ΔV is the potential difference. The formula W = qΔV states that the work done (W) by an electric field to move a charge (q) is equal to the product of the charge (q) and the potential difference (ΔV) between the initial and final points. The potential difference (ΔV) is the change in potential energy per unit charge as the charge moves between those points. The potential at point C (V_C) is 0 volts. The potential at point D (V_D) is -2.7 × 10^-6 volts. The charge being moved (q) is 2 μC, which is equal to 2 × 10^-6 C. Substitute these values into the formula W = qΔV to calculate the work done: - W = (2 × 10^-6 C) × (-2.7 × 10^-6 V) - W = 5.4 × 10^-12 J - Convert joules (J) to microjoules (μJ): - 1 J = 1,000,000 μJ - W = 5.4 × 10^-12 J × (1,000,000 μJ/J) - W = 5.4 μJ The work done by the electric field to move the charge from point C to point D is 5.4 microjoules (μJ). This indicates the amount of energy transferred to the charge by the electric field during its movement. Made with HARPA AI
8:44 al momento de calcular el trabajo , ese trabajo no deberia ir con un menos ? Ya que no es un agente externo quien hace el trabajo si no el propio campo electrico
🎯 Key Takeaways for quick navigation:
The video presents an electrostatics problem involving a triangular arrangement of charges.
The goal is to determine the potential at the vertex (point C) and the potential at a specific point (point D), as well as the work done by the electric field to move a charge from point C to point D.
The points A, B, and C represent the corners of an equilateral triangle with a side length of 100 millimeters.
Charge q1 = +8 μC is located at point B, 100 millimeters to the left of charge q2 = -8 μC.
Point D is located 20 millimeters to the left of charge q2.
The potential at point C is zero volts (V = 0 V).
This is because the net potential at a point due to multiple charges is the sum of the individual potentials due to each charge, and the contributions from q1 and q2 cancel each other out at point C.
The potential at point D is -2.7 × 10^-6 volts (V = -2.7 × 10^-6 V).
This is calculated using the equation V = kq/r, where V is the potential, k is Coulomb's constant, q is the charge, and r is the distance between the charge and the point.
The work done by the electric field to move a charge of 2 μC from point C to point D is 5.4 joules (W = 5.4 J).
This is calculated using the equation W = qΔV, where W is the work, q is the charge, and ΔV is the potential difference.
The potential at point C is zero volts (V = 0 V).
This is because the net potential at a point due to multiple charges is the sum of the individual potentials due to each charge, and the contributions from q1 and q2 cancel each other out at point C.
Electric potential is a scalar quantity that represents the potential energy per unit charge at a given point in an electric field.
It is measured in volts (V).
The potential difference between two points is the change in potential energy per unit charge as a charge moves between those points.
The electric field is a vector quantity that describes the force exerted on a unit charge at a given point.
The resultant potential at point C is zero volts (V = 0 V).
This is because the sum of the individual potentials due to the charges at points A and B is zero.
The potential due to charge q1 at point C is positive, while the potential due to charge q2 at point C is negative.
The magnitudes of these two potentials are equal, resulting in a net potential of zero at point C.
The goal is to determine the potential at point D, which is located 20 millimeters to the left of charge q2.
This involves identifying the distances between the charges and point D, applying the formula for electric potential, and calculating the resultant potential.
The distance between point A and point D is assumed to be 80 millimeters.
This is based on the fact that the triangle is equilateral and the total distance between point A and point C (which is also the distance between point B and point C) is 100 millimeters.
The distance between point B and point D is 20 millimeters.
The formula for electric potential (V) at a point due to a charge (q) is: V = k * q / r, where k is Coulomb's constant (9 × 10^9 N·m^2/C^2) and r is the distance between the charge and the point.
Using this formula, calculate the potential (V_A) at point D due to charge q1 (8 μC) and the potential (V_B) at point D due to charge q2 (-8 μC).
The resultant potential (V_D) at point D is the sum of the potentials (V_A and V_B) due to the individual charges: V_D = V_A + V_B.
Substitute the calculated values for V_A and V_B into the equation to determine the resultant potential at point D.
The goal is to determine the work done by the electric field to move a charge of 2 μC from point C to point D.
This involves applying the formula for work done, which is W = qΔV, where W is the work, q is the charge, and ΔV is the potential difference.
The formula W = qΔV states that the work done (W) by an electric field to move a charge (q) is equal to the product of the charge (q) and the potential difference (ΔV) between the initial and final points.
The potential difference (ΔV) is the change in potential energy per unit charge as the charge moves between those points.
The potential at point C (V_C) is 0 volts.
The potential at point D (V_D) is -2.7 × 10^-6 volts.
The charge being moved (q) is 2 μC, which is equal to 2 × 10^-6 C.
Substitute these values into the formula W = qΔV to calculate the work done:
- W = (2 × 10^-6 C) × (-2.7 × 10^-6 V)
- W = 5.4 × 10^-12 J
- Convert joules (J) to microjoules (μJ):
- 1 J = 1,000,000 μJ
- W = 5.4 × 10^-12 J × (1,000,000 μJ/J)
- W = 5.4 μJ
The work done by the electric field to move the charge from point C to point D is 5.4 microjoules (μJ).
This indicates the amount of energy transferred to the charge by the electric field during its movement.
Made with HARPA AI
8:44 al momento de calcular el trabajo , ese trabajo no deberia ir con un menos ? Ya que no es un agente externo quien hace el trabajo si no el propio campo electrico
Muchísimas gracias le entendí todo.
Me salvó la vida para el examen de mañana, gracias.
una duda, en la pregunta c el potencial no es Vd - Vc por que seria potencial final menos potencial inicial
El ejercicio dice del punto C al D, en ese orden, seria Vc - Vd
Buen video bro
gracias :3
en 6:34 por que ya no va el r al cuadrado
El r² es para Campo Eléctrico (E) o para Fuerza Eléctrica (F), no para potencial eléctrico (V)
en el 7:02 no deberia ser en Vb = -3,6 x 10^5 v?
sería -36x10^5
pero es lo mismo