Respuesta :
The electric force exerted by S1 on S2 is 21.58μN. The electric field created by S1 at the level of S2 is [tex]1.20 \frac{kN}{C}[/tex]. The electric potential created by S1 at the level or S2 is 360V. The electric force exerted by S1 on S2 is 21.58μN. The electric field generated by S1 in the middle of S1 and S2 is [tex]4.79 \frac{kN}{C}[/tex]. The electric field in the middle of S1 and S2 is [tex]11.99 \frac{kN}{C}[/tex]. The electric potential in the middle of S1 and S2 is 1.80 kV. The electric potential generated by S2 on the position of S1 is 539.4V
a) The electric force exerted by S1 on S2 is 21.58μN.
b) The electric field created by S1 at the level of S2 is [tex]1.20 \frac{kN}{C}[/tex]
c) The electric potential created by S1 at the level or S2 is 360V
d) The electric force exerted by S1 on S2 is 21.58μN.
e) The electric field generated by S1 in the middle of S1 and S2 is [tex]4.79 \frac{kN}{C}[/tex]
f) The electric field in the middle of S1 and S2 is [tex]11.99 \frac{kN}{C}[/tex]
g) The electric potential in the middle of S1 and S2 is 1.80 kV
h) The electric potential generated by S2 on the position of S1 is 539.4V
The magnitude of the force is determined by using the following formula:
[tex]F_{e}=k_{e}\frac{|q_{1}||q_{2}|}{r^{2}}[/tex]
where:
F= Electric force [N]
k = Electric constant ([tex]N\frac{m^{2}}{C^{2}}[/tex])
q1= First charge [C]
q2 = Second charge [C]
r = distance between the two charges
So, in this case, the force can be calculated like this:
[tex]F_{e}=(8.99x10^{9}N\frac{m^{2}}{C^{2}})\frac{|12x10^{-9}C||18x10^{-9}C|}{(30x10^{-2}m)^{2}}[/tex]
So the force will be equal to:
[tex]F=21.58x10^{-6}N[/tex]
which is the same as:
[tex]F=21.58 \micro N[/tex]
b) The electric field created by S1 at the level of S2 is [tex]1.20 \frac{kN}{C}[/tex]
We can take the following formula for the electric field.
[tex]E_{S1}=\frac{F_{e}}{q_{2}}[/tex]
[tex]E_{S1}=\frac{1.20 x10^{-6}N}{18x10^{-9}C}[/tex]
which yields:
[tex]E_{S1}=1.20x10^{3} \frac{N}{C}[/tex]
[tex]E_{S1}=1.20 \frac{kN}{C}[/tex]
In this case, since S1 is positive, we can asume the electric field is in a direction away from S1.
c)
The electric potential created by S1 at the level of S2 is 360V
This electric potential can be found by using the following formula:
V=Er
Where V is the electric potential and it is given in volts.
So we can use the data found in the previous sections to find the electric potential:
[tex]V=(1.20x10^{3} \frac{N}{C})(30x10^{-2}m)[/tex]
V=360V
d) The force exerted by S2 on S1 will be the same in magnitude as the force exerted by S1 on S2 but oposite in direction. This is because the force will depend on the two charges, and the distance between them, so:
The electric force exerted by S1 on S2 is 21.58μN.
The magnitude of the force is determined by using the following formula:
[tex]F_{e}=k_{e}\frac{|q_{1}||q_{2}|}{r^{2}}[/tex]
[tex]F_{e}=(8.99x10^{9}N\frac{m^{2}}{C^{2}})\frac{|12x10^{-9}C||18x10^{-9}C|}{(30x10^{-2}m)^{2}}[/tex]
So the force will be equal to:
[tex]F=21.58x10^{-6}N[/tex]
which is the same as:
[tex]F=21.58 \micro N[/tex]
e) The electric field generated by S1 in the middle of S1 and S2 is [tex]4.79 \frac{kN}{C}[/tex]
In order to find the electric field generated by S1, we can make use of the following formula
[tex]E=k_{e} \frac{q_{1}}{r_{1}^{2}}[/tex]
[tex]E=(8.99x10^{9} N\frac{m^{2}}{C^{2}})(\frac{12x10^{-9}C}{(15x10^{-2}m)^{2}})[/tex]
which yields:
[tex]E=4.79 \frac{kN}{C}[/tex]
f) The electric field in the middle of S1 and S2 is [tex]11.99 \frac{kN}{C}[/tex]
In order to find the electric field generated by two different charges at a given point is found by using the following formula:
[tex]E=k_{e} \sum \frac{q_{i}}{r_{i}^{2}}[/tex]
where:
[tex]q_{i}[/tex]= each of the charges in the system
[tex]r_{i}[/tex]= the distance between each of the charges and the point we are analyzing.
We'll suppose the electric fields will be positive then, so:
[tex]E=(8.99x10^{9} N\frac{m^{2}}{C^{2}})[\frac{12x10^{-9}C}{(15x10^{-2}m)^{2}}+\frac{18x10^{-9}C}{(15x10^{-2}m)^{2}}][/tex]
which yields:
[tex]E=11.99 \frac{kN}{C}[/tex]
g) The electric potential in the middle of S1 and S2 is 1.80 kV
Since we know what the electric field is from the previous question, we can make use of the same formula we used before to find the electric potential in the middle of S1 and S2
So let's take the formula:
V=Er
So we can use the data found in the previous sections to find the electric potential:
[tex]V=(11.99x10^{3} \frac{N}{C})(15x10^{-2}m)[/tex]
V=1.80kV
h)
The electric potential generated by S2 on the position of S1 is 539.4V and can be found by using the following formula:
[tex]V=k_{e}\frac{q_{2}}{r}[/tex]
So we can use the data provided by the problem to find the electric potential.
[tex]V=(8.99x10^{9} N\frac{m^{2}}{C^{2}})(\frac{18x10^{-9}C}{30x10^{-2}m})[/tex]
V=539.4V
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1) F12 = 216* 10⁻⁷ [N]
b) E1 = 12*10⁻² [ N / C]
c) V(s1/s2) = 3.6 [Volts]
d)|F21| = 216* 10⁻⁷ [N] For direction of F21 seeAnnex
e)Es₁/p = 0,48*10⁴ [ N / C]
f) Et = 1.2 *10⁴ [N/C]
g)V(p) = 1800 [V]
h) V(s2/s1) = 540 [V]
The electric force due to the small sphere S1 over S2, is equal in module and of opposite direction at the force that S2 produces on S1. As the charges on both spheres are of opposite signs they attract each other.
See Annex: Notice directions of F1 and F2
Let´s call F1 the force exerted by S1 on S2 then according to Coulomb´s law
F1 = K * Q*q/d²
In that equation, K is the constant of Coulomb K = 9*10⁹ N-m²/C²
Q = 12 nC = 12 * 10⁻⁹ C
q = - 18 nC = - 18 *10⁻⁹ C
d = 30 cm = 0,3 m ⇒ d² = 0.09 m²
By substitution we get:
F12 = 9*10⁹ * 12 * 10⁻⁹ *18 *10⁻⁹ / 0.09 N-m²/C² *C*C/m²
F12 = 9*12*18*10⁻⁹/ 0.09 [N]
F12 = 216* 10⁻⁷ [N]
According to what was explained at the beginning F2 ( the electric force exerted by S2 on S1 is equal in module and of opposite direction ( see Annex)
b) The electric field created by S1 at the level of S2
Let´s call that electric field E1
E1 = K * Q / d²
Then
E1 = 9*10⁹* 12 * 10⁻⁹/0.09 [ N-m²/C²*C/m²]
E1 = 12*10⁻² [ N / C]
c) Electric potential is created by S1 at S2 level
V = K * Q/d
V(s1/s2) = 9*10⁹ * 12 * 10⁻⁹/ 0,3 [ N* m²/C² * C /m]
V(s1/s2) = 3*12 *10⁻¹ [N*m/C] ⇒ V(s1/s2) = 3.6 [ J/C ] ⇒
V(s1/s2) = 3.6 [Volts]
d) Already explain in 1 then
|F21| = |F12|
|F21| = 216* 10⁻⁷ [N] For direction of F21 seeAnnex
e)
Es₁/p = K * Q / d²
Es₁/p = 9*10⁹* 12 * 10⁻⁹/(0.15)²
Es₁/p = 9*10⁹* 12 * 10⁻⁹ / 0.0225
Es₁/p = 108 * 10⁴ / 225 [ N / C]
Es₁/p = 0,48*10⁴ [ N / C]
f) To determine the electric field in the middle ( point P)
We need to calculate the electric field due to charge S2 ( - 18 *10⁻⁹ C)
Then
Es2/p = K *q / d²
Es2/p = 9*10⁹* ( - 18 *10⁻⁹ )/ (0.15)²
Es2/p = - 162* 10⁴ /225
Es2/p = - 0.72 * 10⁴ [ N / C]
The electric field ( a vector ) and the force due to a charge ( a vector ) both have the same direction.
Assuming we place a test charge (+) in the middle the sphere S1 will reject such a charge, and S2will attract the test charge therefore the resultant field is the sum of both components, then:
Et = Es₁/p + Es2/p
Et = 0,48*10⁴ + 0,72*10⁴
Et = 1.2 *10⁴ [N/C]
g) To calculate the electric potential at p.
Vp = V s1/p + V s2/p
V s1/p = K * Q /d/2
V s1/p = 9 * 10⁹ * 12*10⁻⁹ / 0,15
V s1/p = 720 [V]
V s2/p = 9 * 10⁹ * 18*10⁻⁹ /0,15
V s2/p = 1080 [V]
Then
V(p) = 1080 + 720
V(p) = 1800 [V]
h) The electric potential at S1 level is due to charge in S2
V(s2/s1) = K * q / d
V(s2/s1) = 9*10⁹ * 18 * 10⁻⁹ / 0,3
V(s2/s1) = 540 [V]
