Electric field and electric potential exercise

  • #1
Homework Statement:
Consider a very large flat surface located in the plane [itex]z = 0[/itex] that is uniformly loaded with a density [itex]\sigma =3\, \textrm{nC/m}^2[/itex]. Consider also a point charge [itex]q=100\, \textrm{pC}[/itex] located at point [itex](0,0,20)\, \textrm{cm}[/itex].
a) What force acts on the charge [itex]q[/itex]?
b) What is the flux of the electric field through a spherical surface of [itex]10\, \textrm{cm}[/itex]-radius centered on the coordinate origin?
c) What force would act on a point charge [itex]Q=-200\, \textrm{pC}[/itex] if we placed it at point [itex](0,5,20)\, \textrm{cm}[/itex]?
d) In the situation of the previous section, what would be the flux of the electric field through a spherical surface of [itex]10\, \textrm{cm}[/itex] radius centered on the charge [itex]Q[/itex]?
Relevant Equations:
[tex] \vec{F}=\vec{E}\cdot q[/tex]
[tex] \phi =\oint \vec{E}d\vec{S}[/tex]
a)
D3F6529A-2AA2-4BB2-B896-1E8D96693A08.jpeg

[tex] \vec{F}=\vec{E}\cdot q [/tex]
[tex] \phi =\oint \vec{E}d\vec{S}=\oint \vec{E}d\vec{S}=\underbrace{\oint \vec{E}d\vec{S}}_{\textrm{FACES } \perp}+\underbrace{\oint \vec{E}d\vec{S}}_{\textrm{FACES } \parallel}=0+\oint EdS\cdot \underbrace{\cos 0}_1= E2S[/tex]
[tex] \dfrac{Q_{enc}}{\varepsilon_0}=\phi [/tex]
[tex]
\left.
\dfrac{Q_{enc}}{\varepsilon_0}=E2S \atop
\sigma =\dfrac{Q_{enc}}{S}
\right\} \quad \sigma S=E2S\varepsilon_0 \rightarrow E=\dfrac{\sigma}{2\varepsilon_0}=\dfrac{3\cdot 10^{-9}}{2\cdot 8'85\cdot 10^{-12}}=169'49\, \widehat{k}\, \textrm{N/C}
[/tex]
[tex] \vec{F}=\dfrac{\sigma}{2\varepsilon_0}\cdot q=\dfrac{3\cdot 10^{-9}}{2\cdot 8'85\cdot 10^{-12}}\cdot 1\cdot 10^{-10}=1'69\cdot 10^{-8}\, \textrm{N}\rightarrow \boxed{\vec{F}=17\widehat{k}\, \textrm{nN}} [/tex]

b)
E1B4E73D-5773-4FAD-BEDD-A821A520A349.jpeg

[tex] \phi =\dfrac{Q_{enc}}{\varepsilon_0}=\dfrac{\pi \cdot R^2\sigma}{\varepsilon_0}=\dfrac{\pi \cdot 0'1^2\cdot 3\cdot 10^{-9}}{8'85\cdot 10^{-12}}=\boxed{10'65\, \textrm{Nm}^2/\textrm{C}} [/tex]

c) [tex] Q=-200\, \textrm{pC}=-2\cdot 10^{-10}\, \textrm{C}\quad P_Q=(0,0'05,0'2)\, \textrm{m}\quad P_q=(0,0,0'2)\, \textrm{m} [/tex]
[tex] \vec{E}=169'49\, \widehat{k} [/tex]
[tex] \vec{F_E}=Q\cdot E=-2\cdot 10^{-10}\cdot 169'49\, \widehat{k}=-3'39\cdot 10^{-8}\, \textrm{C}\, \widehat{k} [/tex]
[tex] \vec{r_{Q\to q}}=(0,0'05,0'2)-(0,0,0'2)=(0,0'05,0) [/tex]
[tex] \vec{F_q}=k\cdot \dfrac{Q\cdot q}{r^3}\vec{r_{Q\to q}}=9\cdot 10^9\cdot \dfrac{(-2\cdot 10^{-10})\cdot 1\cdot 10^{-10}}{0'05^3}\cdot (0,0'05,0)=-7'2\cdot 10^{-8}\, \textrm{C}\, \widehat{j} [/tex]
[tex] \boxed{\vec{F}=(-72\, \widehat{j}-34\, \widehat{k})\, \textrm{nN}} [/tex]

d) [tex]
\left.
\vec{E}=169'49\, \widehat{k} \atop
\vec{E_q}=k\cdot \dfrac{q}{r^3}\vec{r_{Q\to q}}=9\cdot 10^9\cdot \dfrac{1\cdot 10^{-10}}{0'05^3}(0,0'05,0)=360\, \widehat{j}
\right\} \vec{E_T}=(360\, \widehat{j}+169'49\, \widehat{k})\, \textrm{N/C}
[/tex]





How would section (d) of this question be calculated?
In section (b) I have applied Gauss law since I could calculate the internal charge of the spherical surface.
It is the one I've made so far. The problem I see is that I can't apply Gauss. But I also don't know the relationship between the direction of the surface differential and the direction of the electric field.

Thanks!
 

Answers and Replies

  • #2
collinsmark
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Homework Statement:: Consider a very large flat surface located in the plane [itex]z = 0[/itex] that is uniformly loaded with a density [itex]\sigma =3\, \textrm{nC/m}^2[/itex]. Consider also a point charge [itex]q=100\, \textrm{pC}[/itex] located at point [itex](0,0,20)\, \textrm{cm}[/itex].
a) What force acts on the charge [itex]q[/itex]?
b) What is the flux of the electric field through a spherical surface of [itex]10\, \textrm{cm}[/itex]-radius centered on the coordinate origin?
c) What force would act on a point charge [itex]Q=-200\, \textrm{pC}[/itex] if we placed it at point [itex](0,5,20)\, \textrm{cm}[/itex]?
d) In the situation of the previous section, what would be the flux of the electric field through a spherical surface of [itex]10\, \textrm{cm}[/itex] radius centered on the charge [itex]Q[/itex]?
Relevant Equations:: [tex] \vec{F}=\vec{E}\cdot q[/tex]
[tex] \phi =\oint \vec{E}d\vec{S}[/tex]

a) View attachment 262684
[tex] \vec{F}=\vec{E}\cdot q [/tex]
[tex] \phi =\oint \vec{E}d\vec{S}=\oint \vec{E}d\vec{S}=\underbrace{\oint \vec{E}d\vec{S}}_{\textrm{FACES } \perp}+\underbrace{\oint \vec{E}d\vec{S}}_{\textrm{FACES } \parallel}=0+\oint EdS\cdot \underbrace{\cos 0}_1= E2S[/tex]
[tex] \dfrac{Q_{enc}}{\varepsilon_0}=\phi [/tex]
[tex]
\left.
\dfrac{Q_{enc}}{\varepsilon_0}=E2S \atop
\sigma =\dfrac{Q_{enc}}{S}
\right\} \quad \sigma S=E2S\varepsilon_0 \rightarrow E=\dfrac{\sigma}{2\varepsilon_0}=\dfrac{3\cdot 10^{-9}}{2\cdot 8'85\cdot 10^{-12}}=169'49\, \widehat{k}\, \textrm{N/C}
[/tex]
[tex] \vec{F}=\dfrac{\sigma}{2\varepsilon_0}\cdot q=\dfrac{3\cdot 10^{-9}}{2\cdot 8'85\cdot 10^{-12}}\cdot 1\cdot 10^{-10}=1'69\cdot 10^{-8}\, \textrm{N}\rightarrow \boxed{\vec{F}=17\widehat{k}\, \textrm{nN}} [/tex]

b) View attachment 262685
[tex] \phi =\dfrac{Q_{enc}}{\varepsilon_0}=\dfrac{\pi \cdot R^2\sigma}{\varepsilon_0}=\dfrac{\pi \cdot 0'1^2\cdot 3\cdot 10^{-9}}{8'85\cdot 10^{-12}}=\boxed{10'65\, \textrm{Nm}^2/\textrm{C}} [/tex]

c) [tex] Q=-200\, \textrm{pC}=-2\cdot 10^{-10}\, \textrm{C}\quad P_Q=(0,0'05,0'2)\, \textrm{m}\quad P_q=(0,0,0'2)\, \textrm{m} [/tex]
[tex] \vec{E}=169'49\, \widehat{k} [/tex]
[tex] \vec{F_E}=Q\cdot E=-2\cdot 10^{-10}\cdot 169'49\, \widehat{k}=-3'39\cdot 10^{-8}\, \textrm{C}\, \widehat{k} [/tex]
[tex] \vec{r_{Q\to q}}=(0,0'05,0'2)-(0,0,0'2)=(0,0'05,0) [/tex]
[tex] \vec{F_q}=k\cdot \dfrac{Q\cdot q}{r^3}\vec{r_{Q\to q}}=9\cdot 10^9\cdot \dfrac{(-2\cdot 10^{-10})\cdot 1\cdot 10^{-10}}{0'05^3}\cdot (0,0'05,0)=-7'2\cdot 10^{-8}\, \textrm{C}\, \widehat{j} [/tex]
[tex] \boxed{\vec{F}=(-72\, \widehat{j}-34\, \widehat{k})\, \textrm{nN}} [/tex]

d) [tex]
\left.
\vec{E}=169'49\, \widehat{k} \atop
\vec{E_q}=k\cdot \dfrac{q}{r^3}\vec{r_{Q\to q}}=9\cdot 10^9\cdot \dfrac{1\cdot 10^{-10}}{0'05^3}(0,0'05,0)=360\, \widehat{j}
\right\} \vec{E_T}=(360\, \widehat{j}+169'49\, \widehat{k})\, \textrm{N/C}
[/tex]





How would section (d) of this question be calculated?
In section (b) I have applied Gauss law since I could calculate the internal charge of the spherical surface.
It is the one I've made so far. The problem I see is that I can't apply Gauss. But I also don't know the relationship between the direction of the surface differential and the direction of the electric field.

Thanks!
Parts a) through c) look good to me. :smile:

For part d) you can apply Gauss' Law just fine. The problem statement isn't asking for [itex] \vec E [/itex], it's asking for the flux [itex] \phi [/itex].

You are correct that Gauss's Law is not very useful in finding [itex] \vec E [/itex], except in certain situations where symmetry is present, such as a spherically symetrical charge, an infinitely long wire with cylindical charge symmetry, or an infinite plane with uniform charge -- and all assuming that there are no other charges around. That's because only in those special cases is [itex] |\vec E| [/itex] constant. So for the situation described in part d), you won't be able rely on Guass' Law to simply derive [itex] \vec E [/itex]. But part d) of this problem is not asking you to find [itex] \vec E [/itex].

This problem (part d) is asking you to find the flux. Gauss' Law always holds true whether or not [itex] E [/itex] across the surface is constant or not. The total flux is proportional to the charge enclosed. Always. [Edit: just make sure to properly account for all the charge enclosed within the surface.]

(Gauss's law is one of Maxwell's Equations, after all. And in classical electrodynamics it always holds true. Always.)
 
Last edited:
  • #3
Parts a) through c) look good to me. :smile:

For part d) you can apply Gauss' Law just fine. The problem statement isn't asking for [itex] \vec E [/itex], it's asking for the flux [itex] \phi [/itex].

You are correct that Gauss's Law is not very useful in finding [itex] \vec E [/itex], except in certain situations where symmetry is present, such as a spherically symetrical charge, an infinitely long wire with cylindical charge symmetry, or an infinite plane with uniform charge -- and all assuming that there are no other charges around. That's because only in those special cases is [itex] |\vec E| [/itex] constant. So for the situation described in part d), you won't be able rely on Guass' Law to simply derive [itex] \vec E [/itex]. But part d) of this problem is not asking you to find [itex] \vec E [/itex].

This problem (part d) is asking you to find the flux. Gauss' Law always holds true whether or not [itex] E [/itex] across the surface is constant or not. The total flux is proportional to the charge enclosed. Always. [Edit: just make sure to properly account for all the charge enclosed within the surface.]

(Gauss's law is one of Maxwell's Equations, after all. And in classical electrodynamics it always holds true. Always.)
Aa okay, that is even if it is not constant you know that it will surely be proportional because it must always be a multiple of the elementary charge, the electron. Right?
 
  • #4
Delta2
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Aa okay, that is even if it is not constant you know that it will surely be proportional because it must always be a multiple of the elementary charge, the electron. Right?
Not sure what you trying to say here, but I think @collinsmark meant that when symmetry is present, ##|\vec{E}|## is constant throughout the surface in which we calculate the integral of gauss's law.

For example we have a point charge q . Lets choose as gaussian surface a spherical surface of radius ##R##, centered at the point charge. Then the magnitude of the electric field throughout this spherical surface is constant and equal to ##k\frac{q}{R^2}##. It isn't constant in all the space around the point charge.
 
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  • #5
collinsmark
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Aa okay, that is even if it is not constant you know that it will surely be proportional because it must always be a multiple of the elementary charge, the electron. Right?
It doesn't have to do with a multiple of the elementary charge per se (not in classical electrodynamics anyway). Gauss' Law was around long before the electron was discovered.

Before you answer part d), it might help to put Gauss' Law into words. Give it a try, but to do so by keeping it simple, without using too many mathematics terms. Try not to use words like "integral." And don't worry much about [itex] \varepsilon_0 [/itex] -- that's just a constant of proportionality and is not of particular importance for this.

[tex] \oint \vec E \cdot \vec {dS} = \frac{Q_{enc}}{\varepsilon_0} [/tex]

Also, you know that (using the above notation) that for the closed surface [itex] S [/itex], the flux is [itex] \phi = \oint \vec E \cdot \vec {dS} [/itex]. Also note that [itex] S [/itex] can be any closed surface. It doesn't have to be a sphere, or a cylinder, or pillbox. It can be anything. It can even be in the shape of some weird amorphous blob, so long as it is a closed surface.

So in simple terms, what is Gauss' Law really saying?

[Edit: it's totally ok here to use terms like "electric field lines," or "lines of flux," or somesuch, if you wish.]
 
Last edited:
  • #6
For example we have a point charge q . Lets choose as gaussian surface a spherical surface of radius ##R##, centered at the point charge. Then the magnitude of the electric field throughout this spherical surface is constant and equal to ##k\frac{q}{R^2}##. It isn't constant in all the space around the point charge.
Exactly, perfect.
 
  • #7
Before you answer part d), it might help to put Gauss' Law into words. Give it a try, but to do so by keeping it simple, without using too many mathematics terms. Try not to use words like "integral." And don't worry much about [itex] \varepsilon_0 [/itex] -- that's just a constant of proportionality and is not of particular importance for this.
Okay, perfect.
So in simple terms, what is Gauss' Law really saying?
Gauss law would say that the charge inside a closed surface (which can take any shape) is proportional to the flux through it, wouldn't it?
 
  • #8
collinsmark
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Okay, perfect.

Gauss law would say that the charge inside a closed surface (which can take any shape) is proportional to the flux through it, wouldn't it?
Yes! :biggrin:

And there are some other things that go with this. One of which we already established:
  • It doesn't say anything about the shape of the surface. The total flux through the closed surface is only depended upon the charge within that surface. The shape of the surface is irrelevant, so long as the intricacies of the surface's shape do not effect the total charge enclosed within.
  • Nothing outside the surface affects the total flux through the surface. When determining the total flux through the closed surface, the only charges that matter are those within the surface. Those things outside the surface don't matter.
  • For those charges that are inside the closed surface, it doesn't matter where they are inside the surface when determining the total flux through the surface. You can move charges around inside the surface all day, so long as no charges go from inside to outside the surface (or vice-versa); but the total flux through the surface will not change.
You should now have all you need to complete part d)*. :smile:

*(For future reference, keep in mind that Gauss's Law only makes claims about the total flux through a closed surface. Moving charges around inside the surface or outside the surface, can change the electric field and flux through small sections of the surface -- it's just that they don't effect the total flux through the closed surface. E.g., moving a charge within the surface will increase the flux in some part of the surface, but it also will decrease the flux in other parts of the surface, and thus the total flux through the overall closed doesn't change. That's Gauss' Law.)
 
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