A student throws a 140 g snowball at 6.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of
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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
