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
Answer:
Required charge .
Explanation:
Given:
Diameter of the isolated plastic sphere = 25.0 cm
Magnitude of the Electric field = 1500 N/C
now
Electric field (E) is given as:
where,
k = coulomb's constant = 9 × 10⁹ N
q = required charge
r = distance of the point from the charge where electric field is being measured
The value of r at the just outside of the sphere =
thus, according to the given data
or
or
Required charge .
Now,
the number of electrons (n) required will be
or
or
The wavelength of the incident photon is .
What is wavelength?
The wavelength, or the distance over which the shape of a periodic wave repeats, is the spatial period in physics. It is a property of both traveling waves and standing waves, as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two neighboring crests, troughs, or zero crossings. The Greek letter lambda is frequently used to denote wavelength. The term wavelength is also sometimes used to describe modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids.
The relationship between wavelength and frequency is inverse, assuming a sinusoidal wave flowing at a constant speed.
Calculations:
The energy loss Δλ=h/(1-cos∅)
Conservation of momentum gives,
Wavelength(λ)==
Wavelength(λ)=
To learn more about wavelength , visit:
#SPJ4