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

[@TheOrganicChemistryTutor]

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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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[@daniellaughoc4003]

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

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Clearest explanation

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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.

[@tagekumarpaul7474]

No matter where you draw the line from the ring to point p R will be the same value.

[@regnas7868]

I figured it out after looking at it again, the problem was that i was looking at it in 2 dimension. However, in a 3 dimension space, the point P is above the circle which will give a right triangle in all cases. It was just a issue of imagination lol

[@tagekumarpaul7474]

Haha I was struggling with the calculus portion of this too. Cheers mate

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[@alexmichaelsafayan1814]

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

[@mobilegameplay2090]

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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[@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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That's help me soo much but I have question in the Last equation we got if x=0,x>>0, x

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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?

[@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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[@v.m.5850]

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

[@sunildhaul2397]

You could do that also and still you will get the same answer. In place of dQ you would use charge per unit length X dl where dl is an infinitesimal length along the circumference. You would then get E net in terms of linear charge density.

[@daniellaughoc4003]

@@sunildhaul2397 How is this different from the one with the formula of the denominator raised to 3/2……?

[@sunildhaul2397]

@@daniellaughoc4003 , it will ultimately give the same end result except that now your final formula obtained after integration would be having linear charge density in it.

[@loucimouadi3983]

I don't understand why he didn't integrate directly dE?

[@gnyszbr4187]

@@sunildhaul2397 will that be the same for all uniformly distributed charges? Is it ok if I do not use density and just integrate dq for those type of questions but with different shapes with uniformly distributed charges?

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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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[@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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