Give step-by-step solution with explanation and final answer:4. Consider a thin, circular disk with radius R and total charge Q distributed uniformly across its
surface. Suppose the disk lies in the

Question

Give step-by-step solution with explanation and final answer:4. Consider a thin, circular disk with radius R and total charge Q distributed uniformly across its
surface. Suppose the disk lies in the zy-plane with its centre at the origin. The electric field
magnitude in the z-direction, E., at a distance d from the origin along the z-axis is given by

2kQ 1
Bd) = 2 | 1- JE
1+—
d?
where k is Coulomb's constant.
(a) Use the binomial approximation to approximate the electric field strength assuming d > R
(i.e. at a point “not too close” to the disk).
(b) The physical interpretation of the result to part (a) is that at lowest-order, the disk looks like
a point charge Q a distance d away. If d = 10R, then use Taylor's inequality to determine
5
an upper bound on the error of such an approximation as a multiple of Ls

VIPSolutions · Accepted Answer

Let's solve both parts step by step:

---

## **(a) Binomial Approximation for $ d \gg R $**

Given:
$$
E_z(d) = \frac{2kQ}{R^2} \left( 1 - \frac{1}{\sqrt{1 + \frac{R^2}{d^2}}} \right)
$$

For $ d \gg R $, $ \frac{R^2}{d^2} \ll 1 $.  
Let's use the binomial approximation for $ (1 + x)^{-1/2} $ where $ x = \frac{R^2}{d^2} $:

$$
(1 + x)^{-1/2} \approx 1 - \frac{1}{2}x + \frac{3}{8}x^2
$$

So,
$$
\frac{1}{\sqrt{1 + \frac{R^2}{d^2}}} \approx 1 - \frac{1}{2}\left(\frac{R^2}{d^2}\right)
$$

Plug this into $ E_z(d) $:

$$
E_z(d) \approx \frac{2kQ}{R^2} \left[ 1 - \left( 1 - \frac{1}{2}\frac{R^2}{d^2} \right) \right]
= \frac{2kQ}{R^2} \left( \frac{1}{2}\frac{R^2}{d^2} \right)
= \frac{kQ}{d^2}
$$

**Explanation:**  
At large distances, the electric field of the disk looks just like that of a point charge $ Q $ at the origin.

---

## **(b) Error Bound Using Taylor's Inequality**

For $ d = 10R $, $ x = \frac{R^2}{d^2} = \frac{1}{100} $.

The next term in the binomial expansion is:
$$
\frac{3}{8}x^2 = \frac{3}{8} \left( \frac{R^2}{d^2} \right)^2
$$

Taylor's inequality gives the error bound as the absolute value of the next term:

$$
\left| \text{Error} \right| \leq \frac{2kQ}{R^2} \cdot \frac{3}{8} \left( \frac{R^2}{d^2} \right)^2
= \frac{2kQ}{R^2} \cdot \frac{3}{8} \cdot \frac{R^4}{d^4}
= \frac{3kQ}{4} \cdot \frac{R^2}{d^4}
$$

Express this as a multiple of $ \frac{kQ}{d^2} $:

$$
\frac{3kQ}{4} \cdot \frac{R^2}{d^4} = \frac{3}{4} \frac{kQ}{d^2} \left( \frac{R^2}{d^2} \right)
$$

For $ d = 10R $:

$$
\frac{R^2}{d^2} = \frac{1}{100}
$$

So, the error bound is:

$$
\left| \text{Error} \right| \leq \frac{3}{4} \frac{kQ}{d^2} \left( \frac{1}{100} \right) = \frac{3}{400} \frac{kQ}{d^2}
$$

---

## **Final Answers**

### **(a)**
$$
E_z(d) \approx \frac{kQ}{d^2}
$$

### **(b)**
For $ d = 10R $, the upper bound on the error is:
$$
\boxed{
\left| \text{Error} \right| \leq \frac{3}{400} \frac{kQ}{d^2}
}
$$

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