Electric Field Due to a Ring of Charge, Linear Charge Density, Physics Practice Problems

This physics video tutorial explains how to calculate the electric field of a ring of charge. It explains why the y components of the electric field cancels and how to calculate the linear charge density given the total charge of the ring, the radius, and the distance between the center of the ring and a point P that lies on the x axis perpendicular to the surface of the ring. This video contains plenty of examples and practice problems.

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[@denizsoydan8235]

But.. didn't we ignore the "+c" for the indefinite integral? Was it somehow definite, did I miss it? Did we say from 0 to Q or something like that?

[@sunildhaul2397]

You're correct. No definite integral was taken therefore some constant c should come into the picture. We can then determine c by substituting E=0 when Q=0, which gives us c=0. Note that after indefinite integration, we get E as a function of Q i.e E = f(Q) + c.
Another way of doing this is to look at an indefinite integral as always being a sum of all infinitesimal parts when applied to a given situation. In this case, we do not need to use definite integral since we know the integral of small E i.e. dE in this situation is the sum of all dEs over the cirumference and the integral of dQ is the sum of all dQs over the cirumference, which is clearly E (net electric field) and Q (total charge on ring) respectively. So, either use definite integral or indefinite integral sum approach.

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this is way nicer than the line of charge problem, atleast you dont have to do a crazy trigometric subsitution

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How is this different from the one with the formula of the denominator raised to 3/2……?

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[@tobiaschapinda6771]

A thin circular ring, of radius 20cm, is charged, with a uniform charge density(rho). If a small section, of 1cm length, is removed from the ring. Calculate the electric field intensity at the center of the ring.

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[@regnas7868]

I don't know why R doesn't change?? I understand that R is in terms of a and x which they are constant; However, if dq will be in another position on the ring, u will not have a right triangle to use the Pythagorean theorem to find R in terms of X and a; therefore, R is not constant. Please correct me. Thank you.

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[@lefterisspanoudakis3152]

How can we prove using math, and not symmetry, that the y-components are add up to zero? Can we take the integral of dE_y from 0 to 2πα or we must calculate two or more integrals because the sign of the y-component is changing?

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[@elvisvasquez7553]

One thing I never understood is why is λ ds = dQ. If λ is already the amount of charge per unit of length, why do we need to multiply it by ds?

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[@arinzeanthony7447]

Please when is E=p/£and when is it p/£.
Where E= Electric field
P= charge density
£= permittivity of free space.

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[@alexmichaelsafayan1814]

How is R a constant? 7:00 - the logic here does not make sense

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Yep I did not get it as well

[@clioosesli4646]

Because the distance from any point of the ring to the point where we are finding the electric field won't change that's why R is a constant

[@alexsafayan7684]

@@clioosesli4646 but that is simply not true… take, for example, the left most point on the ring vs the right most point on the ring. the distance from each of those points to the point of interest is different

[@clioosesli4646]

@@alexsafayan7684 how ?
Okay it's a ring right?
Okay just take a ring shaped bangle or anything and a wooden stick or a pen then pass it through the centre of the ring then you will see that no matter it's from the right left or top bottom or any point the distance is same

[@alexmichaelsafayan1814]

@@clioosesli4646 Oh wow, I think I've been interpreting this as a 2D problem, when in reality, the ring is "facing" the point of interest, correct?

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[@v.m.5850]

Why isn’t the linear charge density inserted before integrating?

[@troymills2614]

Why wasn't the linear charge density inserted into the point charge equation before integration, after pulling out the constants you should have been integrating d(theta) = theta. Which would have cancelled out the 2pi leaving only KQ/(x^2+a^2)^3/2 = 2.05*10^5 N/C !!

[@NikkaPleeease]

The denominator should be (x^2+a^2)^3/2, not R^3.... I think you're right!

[@v.m.5850]

I have the same question, I know this is late but have you figured out why?

[@daniellaughoc4003]

@@NikkaPleeease How is this different from the one with the formula of the denominator raised to 3/2……?

[@Sarah-jr8qv]

@@NikkaPleeease it's the same because R= sqrt of x^2+a^2 which equals to (x^2+a^2)^1/2 so R^3 = (x^2+a^2)^3/2

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