# Stumped on Calculating Total Charge on Conducting Surface

• Dawei
In summary, the conversation is about trying to find the total charge on a conducting surface, but the integral is proving to be difficult. A substitution of u = cos(theta) is suggested to simplify the integral, and the conversation also discusses potential errors and modifications to the integral.
Suppose you make the substitution u = cos(theta). I think that ought to simplify it a bit more.

Edit: I had sin(theta) earlier but I think you want cos.

Yes u = cos theta and du = - sin(theta) d(theta)
yields ... minus Integral (p + qu)^(-3/2) du
where ... p = R^2 + a^2, q = -2Ra
which becomes
- 2 [ (p+qu)^(-1/2) ] + C
backsubstituting
- 2 [ (p+qcos(theta))^(-1/2) ] + C

I think I may be off by a constant factor here.
The (p+qu)^(-3/2) term needs to be modified to
k (p/q +u)^(-3/2) before integrating.

Last edited:
Just because you don't get a response immediately doesn't mean "no one can do it". In fact, that's a fairly elementary integral.

Let $u= R^2+ a^2- 2aR cos(\theta)$ so that $du= 2aR sin(\theta) d\theta$.

When $\theta= 0$, $u= R^2- 2aR+ a^2= (R- a)^2$.

When $\theta= \pi$, $u= R^2+ 2aR+ a^2= (R+ a)^2$.

$$\int_{(R-a)^2}^{(R+a)^2} u^{-3/2} du$$
$$= -\frac{2}{3}u^{-1/2}\right|_{(R- a)^2}^{(R+a)^2}$$
$$= \frac{2}{3}\left(\frac{1}{R-a}- \frac{1}{R+ a}\right)$$
$$= \frac{4}{3}\frac{a}{R^2- a^2}$$

## 1. How do I calculate the total charge on a conducting surface?

To calculate the total charge on a conducting surface, you can use the formula Q = σA, where Q is the total charge, σ is the surface charge density, and A is the area of the surface. Simply plug in the values for σ and A to find the total charge.

## 2. What is surface charge density?

Surface charge density is the amount of charge per unit area on a conducting surface. It is measured in coulombs per square meter (C/m2). This value is important in calculating the total charge on a conducting surface.

## 3. Can the total charge on a conducting surface be negative?

Yes, the total charge on a conducting surface can be negative. This means that there is an excess of negative charge on the surface, which can happen if the surface is in contact with a negatively charged object or if electrons are added to the surface.

## 4. How is the total charge on a conducting surface affected by changes in the surface area?

The total charge on a conducting surface is directly proportional to the surface area. This means that as the surface area increases, the total charge also increases. Similarly, if the surface area decreases, the total charge will decrease.

## 5. What is the difference between conducting and non-conducting surfaces in terms of total charge?

Conducting surfaces allow charges to move freely, so the total charge on a conducting surface will distribute evenly across the entire surface. Non-conducting surfaces, on the other hand, do not allow charges to move, so the total charge will be concentrated in specific areas instead of being evenly distributed.

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