Respuesta :
The definition of electric field, Coulomb's law and Newton's second law allow to find the results for the questions about the electric field and the oscillation of the test charge are:
a) In the attachment you have the scheme of the system
b) The point where the electric field is zero is: x = 0 m
c) The angular velocity of the test charge is w = 0.5 rad / s
Given parameters
- Value of the charges q₁ = q₂ = 7.0 10⁻⁹ C
- Separation of charges d = 1.0 m
- Charge position q₁ is: x₁ = -0.5 m
- The value of the charge 3 is q₃ = 1.00 10⁻⁹C
To find
a) system scheme
b) point where the field is zero
c) The angular velocity of q₃
The electric field is the field produced by a distribution of electric charges at a point in space, it is a vector quantity, for the special case that the charges are in one dimension the sum of the field produced by a series of charges is reduced to the sum algebraic. The field produced by point charges is:
E = [tex]k \sum \frac{q_i}{r_i^2}[/tex]
Where E is the electric field, k is the Coulomb constant, q the value of the charge and r the distances from the charge to the point of interest.
a) In the attachment we can see a diagram of the system, the reference system, the location of the charges and the electric field vectors are shown.
b) From the diagram we see that each field is in the opposite direction, therefore the total field is:
[tex]E_{total} = E_1 - E_2 \\E_{total}= k \frac{q_1}{(-0.5- x)^2} - k \frac{q2}{(0.5-x)^2 }\\\\E_{total = 0[/tex]
[tex]\frac{q}{(0.5+x)^2 } = \frac{q}{(0.5-x)^2}[/tex]
The value of the two charges is the same, therefore the value of position x is:
x = 0 m
The point where the total electric field becomes zero is at the origin of the coordinate system.
c) We look for the value of the force on the test charge introduced q₃, for this using the Coulomb equation that establishes that the force is proportional to the product of the electric charges and is inversely proportional to the square of the distance
[tex]F_{13} = k \frac{q_1 q_3x}{(x+0.5)^2}\\F_{23} = k \frac{q_2 q_3}{(x-0.5)^2 }[/tex]
Newton's second law gives the relationship between the net force and the product of mass and acceleration.
[tex]\sum F = m \a \\k q q_3 [ \frac{1}{(x+0.5)^2} - \frac{1}{(x-0.5)2 } }] = m \frac{d^2x}{dt^2}[/tex]
if we use the binomial development
(0.5 + x)⁻² = 0.5 +2 x
(0.5 -x)⁻² = 0.5 -2x
we substitute
k q q₃ [-4x] = [tex]m \frac{d^2x}{dt^2}[/tex]
[tex]-(\frac{4k \ q \ q_3}{m}) \ x = m \frac{d^2x}{dt^2}[/tex]
In oscillatory motion the general equation of motion is
[tex]- w^2 \ x = \frac{d^2x}{dt^2}[/tex]
Equating the two expressions the angular velocity is
[tex]w^2 = \frac{4k \ q \q_3}{m}[/tex]
Let's calculate.
w²= [tex]\frac{4 \ 9 \ 10^9 \ 7 \ 10^{-9} \ 1 \ 10^{-9}}{1 \ 10^{-6}}[/tex] a 4 9 10⁹ 7 10-9 1 10-9 / 1 10-6
w =[tex]\sqrt{0.252}[/tex]
w = 0.5 rad / s
In conclusion, using the definition of electric field, Coulomb's law and Newton's second law we can find the results for the questions about the electric field and the oscillation of the test charge are:
a) In the attachment you have the scheme of the system.
b) The point where the electric field is zero is: x = 0 m
c) The angular velocity of the test charge is w = 0.5 rad / s
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