Net electric field from multiple charges in 2D | Physics | Khan Academy

In this video David solves an example 2D electric field problem to find the net electric field at a point above two charges. Created by David SantoPietro.

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[aliasanon]

This was a really thorough and detailed explanation that leads to learning

[srenity]

This video was amazing. The small but ubiquitous edits were a great addition to the silky smooth drawing and writing of the work.

[morinkashigon]

Two times learning it from my professor, random youtube video and then I came here to understand this and I finally did.
Thank You Sir.

[jsathibabu]

David sir is special because he follows vector rules for very concept and very problem

[cec]

Wow! you are awesome! you are so thorough !

[niall_al]

thank you so much. Physics is just too difficult to learn by yourself and my lecturer is useless, telling us to look at the textbook which is well over a thousand pages long with overcomplicated explanations for full physics students (I'm just doing a physics module for first year engineers). Thanks so much again 😀

[vincenternst]

Nice, but honestly could have just multiplied Ex with the ratio 3/5 instead of doing the whole trig thing.

[MoogleEJA]

This is wonderful thank you! Reviewing the trig steps as you went a long was really helpful.

[khayalamimahlalela]

you just too good man. I appreciate this

[usermanunknown]

You are doing awesome !! honestly said this is the best explanation I had ever seen

[HenrikMyrhaug]

One thing that annoys me here is that you use the similarity of the triangles to conclude the angles are the same, but then don't do the same to find the ratio between the magnitude of the total field strength and the horizontal component of the field strength. Because the triangle of the total distance from one of the charged particles is a 3-4-5 triangle, and the triangle for the field strength from that charged particle and the distance triangle are similar, that means the field horizontal component of the field strength's magnitude is 3/5 of the magnitude of the field strength.

There was no point in finding the angle and then taking cosine of the angle, since you should know cos (theta) = adjacent/hypothenuse of the field strength triangle = adjacent/hypothenuse of the distance triangle (since the triangles are similar) = 3/5.
So you could have calculated :
2, 88 * 3/5 =1, 73
But you calculated:
2, 88 * cos (arccos (3/5)) = 1, 73
or maybe you used arcsin (4/5) or arctan (4/3) instead of arccos (3/5), but my point is it was unnecessary to find the angle theta.

[littlefighter]

Thanks alot.
Again another video with wonderful quality.
May the lord bless you.
Best wishes from Iran.

[Justtocool]

I kinds wish there was an example of 3 or more charges but besides that, a fantastic explanation

[lead]

Would anyone mind explaining why 2.88 N/C applies to both triangles depicted?

[sarah_garapata]

Is there a video where there is two positive charge and 1 negative charge in an equilateral triangle?

[omarhaitham]

Steps to Solve the 2D Electric Field Problem:

Identify the Charges:

Given two charges, positive (blue) and negative (yellow), determine their magnitudes. In this example, they are both eight nanoCoulombs.
Define the Problem:

Determine the electric field at a specific point, P, which is equidistant from both charges.
Break Down the Problem:

Understand that this is a 2D problem, as opposed to a 1D problem between the charges. The goal is to find the net electric field at point P.
Analyze Electric Fields:

Visualize the electric fields created by each charge at point P. Recognize that they lie in a two-dimensional plane.
Component Analysis:

Break down the electric fields into their horizontal (x) and vertical (y) components.
For each charge, identify the horizontal and vertical components.
Symmetry Consideration:

Observe symmetry in the problem. If charges are equidistant and have the same magnitude, vertical components may cancel out due to symmetry.
Focus on Horizontal Components:

Since the vertical components cancel out, focus on finding the horizontal components of the electric fields.
Magnitude Calculation:

Calculate the magnitude of the electric field using the formula:

Determine the distance (r) between the charge and point P using geometry or the Pythagorean theorem.
Component Direction:

Find the angle of the electric field with respect to the horizontal using trigonometry (e.g., tangent).
Use cosine to determine the horizontal component (Ex) of the electric field.
Repeat for Other Charge:

Repeat the process for the other charge. Note that due to symmetry, the horizontal component of the electric field for the negative charge is the same as that for the positive charge.
Net Electric Field:

Add or subtract the horizontal components of the electric fields based on their directions (right or left).
In this example, since both components point to the right, add them up.
Magnitude of Net Electric Field:

The magnitude of the net electric field is equal to the magnitude of its horizontal component. In cases with only horizontal components, this is straightforward.
Direction of Net Electric Field:

Determine the direction of the net electric field. In this example, it points to the right

[shuvrochamp]

wonderfully explained...but we can do this in a different way as the magnitude of both the charges r the same...so we can put the mathmatical formula of electric dipole...
E=K.p/r^3;p=dipole moment=q*d&r=4m as given...we can use this formula when the point in at the vertical area from the middle point of both charges...then theta=90
the general formula is E=K*p/r^3*(1+3cos^2theta)

[neilmashalkar]

I’m so confused on how you got 58.1 degrees for an angle of a 30-60-90 triangle

[rohop]

Why did u not do 3/5 instead of cos(53.1)

[ahmedsinger]

Me after trying to solve a problem of difficulty 3 in Serway test bank😁