Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a
nd a + 150 nC charge at the point (10.0 cm, 8.00 cm).
(a) Find the total electric force on the + 150-nC charge due to the other two.
(b) What is the electric field at the location of the + 150-nC charge due to the presence of the other two charges?
(a) Find the total electric force on the + 150-nC charge due to the other two.
(b) What is the electric field at the location of the + 150-nC charge due to the presence of the other two charges?
1 answer:
Answer:
(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N
(b) E = -42846.7 N/C
Explanation:
The diagram attached below explains some parameters.
Parameters given:
Charge Q1 = +50 nC at point (0, 0)
Charge Q2 = -50 nC at point (0.1, 0)
Charge Q3 = +150 nC at point (0.1, 0.08)
* The distances are in meters.
(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,
F = F1 + F2
FORCE DUE TO Q1 i.e. F(Q1, Q3)
We have to find the x and y components.
From the diagram, we can find θ using SOHCAHTOA:
θ = tan⁻¹ (0.08/0.1)
θ = 38.66⁰
The distance between Q1 and Q3 can be found using Pythagoras theorem:
x² = 0.08² + 0.1²
x = 0.128 m
F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j
F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ
F(Q1, Q3) = (k * Q1 * Q3) / r²
k = Coulombs constant
F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²
F(Q1, Q3) = 0.00412N
F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66
F1 = 0.00322i + 0.00257j N
FORCE DUE TO Q2 i.e. F(Q2, Q3)
We have to find the x and y components.
F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j
F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0
F(Q2, Q3) = (k * Q2 * Q3) / r²
F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²
F(Q2, Q3) = -0.0105N
F2 = -i0.0105 * cos90 - j0.0105 * cos0
F2 = - 0.0105j N
Hence, the total force will be
F = F1 + F2
F = 0.00322i + 0.00257j - 0.0105j
F = 0.00322i - 0.00793j N
The magnitude of this force is:
|F| = √(0.00322² + (-0.00793²)
|F| = 0.00856N
(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:
E = E1 + E2
E1, electric field due to Q1 = kQ1/r²
E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)
E1 = 27465.8 N/C
E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)
E1 = -70312.5N/C
The total electric field:
E = E1 + E2
E = 27465.8 - 70312.5
E = -42846.7 N/C
You might be interested in
To explain, I will use the equations for kinetic and potential energy:
Potential energy is the potential an object has to move due to gravity. An object can only have potential energy if 1) <u>gravity is present</u> and 2) <u>it is above the ground at height h</u>. If gravity = 0 or height = 0, there is no potential energy. Example:
An object of 5 kg is sitting on a table 5 meters above the ground on earth (g = 9.8 m/s^2). What is the object's gravitational potential energy? <u>(answer: 5*5*9.8 = 245 J</u>)
(gravitational potential energy is potential energy)
<h3>Kinetic energy</h3>Kinetic energy is the energy of an object has while in motion. An object can only have kinetic energy if the object has a non-zero velocity (it is moving and not stationary). An example:
An object of 5 kg is moving at 5 m/s. What is the object's kinetic energy? (<u>answer: 5*5 = 25 J</u>)
<h3>Kinetic and Potential Energy</h3>Sometimes, an object can have both kinetic and potential energy. If an object is moving (kinetic energy) and is above the ground (potential), it will have both. To find the total (mechanical) energy, you can add the kinetic and potential energies together. An example:
An object of 5 kg is moving on a 5 meter table at 10 m/s. What is the objects mechanical (total) energy? (<u>answer: KE = .5(5)(10^2) = 250 J; PE = (5)(9.8)(5) = 245 J; total: 245 + 250 = 495 J</u>)