E-field at the origin from two point charges

[goohu]

Homework Statement

Two point charges Q1 = Q, Q2 = 2Q are located at the points (x1; y1; z1) =
(0; 2; 1), (x2; y2; z2) = (1; 0; 2), determine the electric field at the origin (x; y; z) =
(0; 0; 0).

Relevant Equations

coulomb's law : E = Q/(4*pi*e0*R^2) , R = distance

Using superposition and "breaking up" the vectors into three components ax, ay, az on points should solve the task.

For Q1 there is no effect on x-axis.
On the y-axis the distance from Q1 to origin is 2. Using coulombs law will give us -> (-Q/4) * k , where k is the constant 1/(4*pi*e0).
On the z-axis the distance from Q1 to origin is 1. Using the same calculations we get: -Q*k.

We will use the same calculations for Q2 on all axis so let's me just write the results.
x-axis: distance is 1. We get -2Q*k.
y-axis: no effect.
z-axis: distance is 2. We get -2Q*k * (1/4).

Total electric field on all axis become:
x-axis: -2Q / (4*pi*e0)
y-axis: -Q/4 * 1/(4*pi*e0) = -Q/(16*pi*e0)
z-axis: (-Q - 2Q/4) * k = -3Q/(2*4*pi*e0) = -3Q / (8*pi*e0)

But the answer from the textbook is : Q/(20*√5*pi*e0 ) (-2 ; -2 ; -5 ). Where did I go wrong?

 

[SammyS]

Homework Statement:: Two point charges Q1 = Q, Q2 = 2Q are located at the points (x1; y1; z1) =
(0; 2; 1), (x2; y2; z2) = (1; 0; 2), determine the electric field at the origin (x; y; z) =
(0; 0; 0).
Relevant Equations:: coulomb's law : E = Q/(4*pi*e0*R^2) , R = distance

Using superposition and "breaking up" the vectors into three components ax, ay, az on points should solve the task.

For Q1 there is no effect on x-axis.
On the y-axis the distance from Q1 to origin is 2. Using coulombs law will give us -> (-Q/4) * k , where k is the constant 1/(4*pi*e0).
On the z-axis the distance from Q1 to origin is 1. Using the same calculations we get: -Q*k.

We will use the same calculations for Q2 on all axis so let's me just write the results.
x-axis: distance is 1. We get -2Q*k.
y-axis: no effect.
z-axis: distance is 2. We get -2Q*k * (1/4).

Total electric field on all axis become:
x-axis: -2Q / (4*pi*e0)
y-axis: -Q/4 * 1/(4*pi*e0) = -Q/(16*pi*e0)
z-axis: (-Q - 2Q/4) * k = -3Q/(2*4*pi*e0) = -3Q / (8*pi*e0)

But the answer from the textbook is : Q/(20*√5*pi*e0 ) (-2 ; -2 ; -5 ). Where did I go wrong?

Textbook answer looks good.

How far Q₁ from the origin? (This is not a different number for x, y, and z components.)

How far Q₂ from the origin?

 

[berkeman]

For Q1 there is no effect on x-axis.

What does this mean? Just because the charge is sitting along the x-axis does not mean that it does not have a component of E-field along that axis...

 

[kuruman]

Homework Statement:: Two point charges Q1 = Q, Q2 = 2Q are located at the points (x1; y1; z1) =
(0; 2; 1), (x2; y2; z2) = (1; 0; 2), determine the electric field at the origin (x; y; z) =
(0; 0; 0).
Relevant Equations:: coulomb's law : E = Q/(4*pi*e0*R^2) , R = distance

x-axis: distance is 1. We get -2Q*k.
y-axis: no effect.
z-axis: distance is 2. We get -2Q*k * (1/4).

What do you mean by "distance"?. The electric field at the origin is given by $$\vec E_1=\frac{1}{4\pi \epsilon_0}\frac{(-\vec r_1)}{r_1^{3/2}}.$$Note the denominator. It's not $x_1$ for the x-component, ignored for the y-component and $z_1$ for the z-component. Similarly for charge 2. That' s where you went wrong.

 

[ehild]

Homework Statement:: Two point charges Q1 = Q, Q2 = 2Q are located at the points (x1; y1; z1) =
(0; 2; 1), (x2; y2; z2) = (1; 0; 2), determine the electric field at the origin (x; y; z) =
(0; 0; 0).
Relevant Equations:: coulomb's law : E = Q/(4*pi*e0*R^2) , R = distance

Using superposition and "breaking up" the vectors into three components ax, ay, az on points should solve the task.

you have to use the Coulomb Law in its vector form. The electric field at point $\vec r$ due to a point charge Q located at point $\vec r_1$ is
$\vec E = k\frac{Q}{|\vec r-\vec r_1|^3 }(\vec r-\vec r_1)$

 

[archaic]

For Q1 there is no effect on x-axis.

y-axis: no effect.

If a test charge were to be put on the $x$-axis or the $y$-axis, would you say that it'll feel a force from both charges?

In any case, you have been decomposing the electric field wrongly. In three dimensions, the magnitude of a discrete charge's electric field is given by:
$$|\mathbf E|=\frac{kQ}{x^2+y^2+z^2}$$
Decomposing $|\mathbf E|$ in the $y$ direction, for example, you get:
$$|\mathbf E|_y=|\mathbf E|\cdot\hat y=|\mathbf E|\cos\theta=|\mathbf E|\frac{y}{\sqrt{x^2+y^2+z^2}}=kQ\frac{y}{(x^2+y^2+z^2)^\frac{3}{2}}$$
where $\theta$ is the angle between the $y$-axis and the electric field vector, NOT $\pm\frac{kQ}{y^2}$.
Also, by $x$, $y$ and $z$, it is meant the (signed) distance ($x_\text{point}-x_\text{charge}$ etc...) between the charge and the point in space, so that if your charge is at $(a,\,b,\,c)$ and your variable point is at $(d,\,e,\,f)$, then $x$ wouldn't be just $d$. The same goes for the other coordinates.

 

[ehild]

In any case, you have been decomposing the electric field wrongly. In three dimensions, the magnitude of a discrete charge's electric field is given by:
$$|\mathbf E|=\frac{kQ}{\sqrt{x^2+y^2+z^2}}$$

No, the magnitude of electric field at the origin, due to a point charge Q at point (x,y,z) is
$$|\mathbf E|=\frac{kQ}{x^2+y^2+z^2}$$

 

[archaic]

No, the magnitude of electric field at the origin, due to a point charge Q at point (x,y,z) is
$$|\mathbf E|=\frac{kQ}{x^2+y^2+z^2}$$

$r^2$, my bad.

 

[goohu]

Thanks for the responses everybody. I solved some problems in 2D and mistakenly believed my method was correct.

The distance was kind of confusing me (maybe it still is).

If we look at Q1 for example the distance to origin is √5. It can be verified by pythagoras theorem for two dimensions.

Pythogaras theorem for three dimensions is IIRC : c^2 = x^2 + y^2 + z^2.
If we want the distance squared we should take c^2, so why do we instead calculate r1^(3/2) , which gives the right answer? Don't we want the distance to the point charge squared in the denominator?

 

[kuruman]

Thanks for the responses everybody. I solved some problems in 2D and mistakenly believed my method was correct.

The distance was kind of confusing me (maybe it still is).

If we look at Q1 for example the distance to origin is √5. It can be verified by pythagoras theorem for two dimensions.

Pythogaras theorem for three dimensions is IIRC : c^2 = x^2 + y^2 + z^2.
If we want the distance squared we should take c^2, so why do we instead calculate r1^(3/2) , which gives the right answer? Don't we want the distance to the point charge squared in the denominator?

Look at the numerator. It contains $\vec r-\vec r'$ that provides the direction and also adds an extra dimension of length.

 

[archaic]

Don't we want the distance to the point charge squared in the denominator?

This for calculating the magnitude of the vector, but what you did was decomposing it into its three components and then sum them up using pythagoras' theorem to get the magnitude.
The distance as you are thinking of it is $$\sqrt{(x_\text{point}-x_\text{charge})^2+(y_\text{point}-y_\text{charge})^2+(z_\text{point}-z_\text{charge})^2}$$

 

[ehild]

If we want the distance squared we should take c^2, so why do we instead calculate r1^(3/2) , which gives the right answer? Don't we want the distance to the point charge squared in the denominator?

You need the electric fields of both charges by magnitude and direction.
The magnitude is kQ/r². As for the direction, the electric field vector points from the position of the positive charge to the point where you want to find the field, the origin in this problem. That is, the electric field is parallel to the vector $\vec 0 -\vec r$ To get the electric field vector, you have to multiply the magnitude by the direction vector,
the unit vector parallel to $\vec 0 -\vec r$ which is $$ \frac{\vec 0 -\vec r}{|\vec 0 -\vec r|}$$

 

[ZapperZ]

Thanks for the responses everybody. I solved some problems in 2D and mistakenly believed my method was correct.

The distance was kind of confusing me (maybe it still is).

If we look at Q1 for example the distance to origin is √5. It can be verified by pythagoras theorem for two dimensions.

Pythogaras theorem for three dimensions is IIRC : c^2 = x^2 + y^2 + z^2.
If we want the distance squared we should take c^2, so why do we instead calculate r1^(3/2) , which gives the right answer? Don't we want the distance to the point charge squared in the denominator?

The first thing you should have shown here is a sketch of the problem (which is a requirement for the students in my class).

E₁ is in the yz-plane, while E₂ is in the xz-plane.

Next, are you able to write both E-vectors in their respective components? E₁ only has y and z components, while E₂ only has x and z components.

The resultant will be the sum of each individual components.

At this point, I'm not sure where you are along this step.

Zz.

 

[goohu]

I've the same sketch but it's handwritten and sloppy. Do you recommend any specific tool for those kind of sketches?

Anyway the problem has been solved, so thanks everyone! I used the method as ehild suggested at post number 5.

I suppose the E1 that are shown in the sketch could be written as the sum of the individual components which should be expressed using cos and sin and then divide it by the magnitude.
So for Q1 we get ; E = 2/sqrt(5) at the y direction and 1/sqrt(5) at the x direction. Does this look good?

I do have one last question though:

berkeman wrote;
"Just because the charge is sitting along the x-axis does not mean that it does not have a component of E-field along that axis... "

but ZapperZ then wrote;
"E1 is in the yz-plane, while E2 is in the xz-plane.

Next, are you able to write both E-vectors in their respective components? E1 only has y and z components, while E2 only has x and z components. "

I initially thought that E1 does NOT have a component on the x-axis since it is sitting on the axis. The two posts are giving conflicting information? (English is not my first language.)

 

[etotheipi]

I initially thought that E1 does NOT have a component on the x-axis since it is sitting on the axis. The two posts are giving conflicting information? (English is not my first language.)

The two posts are consistent. $Q_{1}$ is not sitting on the $x$ axis, it is in the $y-z$ plane. You know the field from this charge is going to be parallel to the line joining the charge and the origin, so evidently there can't be any $x$ component!

If the charge were to sit on the $x$ axis (for some reason...), there would only be an $x$ component at the origin.

Like @ZapperZ said, it's really important to consult diagrams to get a hold on the configuration.

 

[goohu]

When you say "sitting on the x-axis" do you mean the position of a charge Q is (x, 0 , 0)?

 

[etotheipi]

When you say "sitting on the x-axis" do you mean the position of a charge Q is (x, 0 , 0)?

Yes. But it was only an example, it's not part of your question.