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Wize University Physics Textbook (Master) > Electrostatic Forces and Electric Fields

Electric Field of Continuous Charge (2D Distributions)

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Example: Electric Field of a Ring


What is the field at point P in the figure below? The total charge on the ring is Q and it has a uniform positive charge distribution.


We need to recognize the symmetry in this situation - the net field is going to be in the x-direction only, pointing to the right. The vertical components are cancelled out because for each piece of ring above the x-axis, there is another piece below the x-axis with an equal contribution - the horizontal components will add together, but the vertical components will be equal and opposite.

Next: what is the x-component of the electric field due to each little piece of the ring?


Each piece of the ring has charge dqdq. We'll write the electric field contribution from one piece of charge, and then substitute rr and θ\theta in terms of the constants aa and LL. (For the angle, note: cos⁡θ=adjacenthypotenuse\cos\theta=\frac{adjacent}{hypotenuse})
dEx=k(dq)r2cos⁡θdEx=k(dq)(a2+L2)2La2+L2dEx=kLdq(a2+L2)32\begin{aligned} dE_x&=\frac{k(dq)}{r^2}\cos\theta \\ dE_x&=\frac{k(dq)}{(\sqrt{a^2+L^2})^2} \frac{L}{\sqrt{a^2+L^2}} \\ dE_x&=\frac{kLdq}{({a^2+L^2})^\frac32} \\ \end{aligned}
Finding the total charge, we need to integrate. This integral is actually really easy, though - the field contribution from each little bit of the ring is the exact same! Everything in the integral is constant and can be pulled out, and we are simply left with ∫dq=Q\int dq = Q.
Ex=∫dExEx=∫kLdq(a2+L2)32Ex=kL(a2+L2)32∫dqEx=kQL(a2+L2)32\begin{aligned} E_x &= \int dE_x \\ E_x &= \int \frac{kLdq}{({a^2+L^2})^\frac32} \\ E_x &= \frac{kL}{({a^2+L^2})^\frac32} \int dq\\ E_x &= \frac{kQL}{({a^2+L^2})^\frac32} \\ \end{aligned}


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Example: Electric Field of a Disk


a) What is the field at point P in the figure below, at some distance L away from a uniformly charged disc of surface charge density σ\sigma?
b) What is the field at point P if the distance L is much greater than the radius of the ring R?
c) What is the field at point P if the radius of the ring R is much greater than the distance L?


Remember that for a ring of radius a and total charge Q (uniformly distributed), the field contribution to the point P was:
Ex=kQL(a2+L2)32E_x = \frac{kQL}{({a^2+L^2})^\frac32} \\
Hint 1: You'll need the following integral:
∫xdx(a2+x2)3/2=−1a2+x2+C\int\frac{xdx}{(a^2+x^2)^{3/2}}=-\frac{1}{\sqrt{a^2+x^2}}+C
Hint 2: You will find the following Taylor approximation useful (valid only for very small xx):
11+x2≈1−12x2\frac{1}{\sqrt{1+x^2}}\approx1-\frac12x^2


Part a)

We're going to treat the disk as a bunch of rings, of radius 0 to R. So we're going to integrate over the "a" variable from before, which was the radius of a single ring.

The charge of each ring, in this case, is the area of the ring times the charge density. The area of a ring is the thickness, da, times the circumference, 2πa2\pi a.

Then the charge of each of our (imaginary) rings is: dq=2πaσdadq=2\pi a \sigma da
In our earlier equation for the ring, "Q" was the total charge on the ring. We substitute our new charge for a ring:
dEx=kL(2πaσda)(a2+L2)32dE_x = \frac{kL(2\pi a \sigma da)}{({a^2+L^2})^\frac32} \\
Integrate over all rings to get the field from the disk:
Ex=∫0RdExEx=∫0RkL2πσada(a2+L2)32Ex=kL2πσ∫0Rada(a2+L2)32Ex=kL2πσ(−1a2+L2)a=0a=REx=kL2πσ(1(0)2+L2−1(R)2+L2)Ex=kL2πσ(1L−1R2+L2)Ex=k2πσ(1−LR2+L2)\begin{aligned} E_x&=\int_0^R dE_x \\ E_x&=\int_0^R \frac{kL2\pi \sigma a da}{({a^2+L^2})^\frac32} \\ E_x&=kL2\pi \sigma \int_0^R \frac{a da}{({a^2+L^2})^\frac32} \\ E_x&=kL2\pi \sigma(-\frac{1}{\sqrt{a^2+L^2}})_{a=0}^{a=R} \\ E_x&=kL2\pi \sigma(\frac{1}{\sqrt{(0)^2+L^2}}-\frac{1}{\sqrt{(R)^2+L^2}}) \\ E_x&=kL2\pi \sigma(\frac 1L-\frac{1}{\sqrt{R^2+L^2}}) \\ E_x&=k2\pi \sigma(1-\frac{L}{\sqrt{R^2+L^2}}) \\ \end{aligned}
Part b)

If we are a very far distance from the disk (L>>RL>>R), we should expect the electric field to reduce the formula for a simple point charge (from very far away, the disk "looks like" a point charge). Indeed, this is what happens:

Ex=k2πσ(1−LR2+L2)Ex=k2πσ(1−LL2((RL)2+1))Ex=k2πσ(1−LL((RL)2+1))Ex=k2πσ(1−1((RL)2+1))Ex≈k2πσ[1−(1−12(RL)2)]Ex≈k2πσ[12(RL)2]Ex≈kπ(QπR2)R2L2Ex≈kQL2\begin{aligned} E_x&=k2\pi \sigma(1-\frac{L}{\sqrt{R^2+L^2}}) \\ E_x&=k2\pi \sigma(1-\frac{L}{\sqrt{L^2((\frac{R}{L})^2+1)}}) \\ E_x&=k2\pi \sigma(1-\frac{L}{L\sqrt{((\frac{R}{L})^2+1)}}) \\ E_x&=k2\pi \sigma(1-\frac{1}{\sqrt{((\frac{R}{L})^2+1)}}) \\ E_x& \approx k2\pi \sigma[1-(1-\frac12(\frac{R}{L})^2)] \\ E_x& \approx k2\pi \sigma[\frac12(\frac{R}{L})^2] \\ E_x& \approx k\pi (\frac{Q}{\pi R^2})\frac{R^2}{L^2} \\ E_x& \approx k\frac{Q}{L^2} \\ \end{aligned}
Remember, L in this case is the distance from the point of interest to the charge distribution (R was the radius of the ring), so this is actually the standard electric field formula again!

Part c)

If the radius of the ring is very large, this is effectively an infinite plate of charge.
Ex=k2πσ(1−LR2+L2)Ex=k2πσ(1−LR2(1+(LR)2))Ex=k2πσ(1−LR(1+(LR)2))Ex≈k2πσ(1−LR1+((0)2))Ex≈k2πσ(1−LR)Ex≈k2πσ(1−0)Ex≈(14πϵo)2πσEx≈σ2ϵo\begin{aligned} E_x&=k2\pi \sigma(1-\frac{L}{\sqrt{R^2+L^2}}) \\ E_x&=k2\pi \sigma(1-\frac{L}{\sqrt{R^2(1+(\frac{L}{R})^2)}}) \\ E_x&=k2\pi \sigma(1-\frac{L}{R\sqrt{(1+(\frac{L}{R})^2)}}) \\ E_x&\approx k2\pi \sigma(1-\frac{L}{R\sqrt{1+((0)^2)}}) \\ E_x&\approx k2\pi \sigma(1-\frac{L}{R}) \\ E_x&\approx k2\pi \sigma(1-0) \\ E_x&\approx (\frac1{4\pi \epsilon_o})2\pi \sigma \\ E_x&\approx \frac{\sigma}{2 \epsilon_o} \\ \end{aligned}
We will recover this result again in the section from Gauss's law.

Practice: Electric Field at the Center of a Half-Ring


What is the field at point P in the figure below? The total charge on the charge distribution is Q and it has a uniform positive charge distribution.

Magnitude