# Derivation of induced charge on a dielectric

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1. Dec 4, 2015

### Titan97

1. The problem statement, all variables and given/known data
Show the the induced charge density on a dielectric placed inside a capacitor is given by $$\frac{k-1}{k}\sigma$$ where $\sigma$ is the charge density of the capacitor plates and $k$ is the dielectric constant.

2. Relevant equations
$$E=\frac{E_0}{k}$$

3. The attempt at a solution

Using the above image, let $E_0$ be the electric field before inserting the dielectric and $E_i$ be the induced electric field.

Total electric field between the plates $$E=E_0-E_i$$
$$\frac{E_0}{k}=E_0-E_i$$
$$E_i=\frac{k-1}{k}E_0$$

Now, $E_0$ is the total electric field due to the plates of the capacitor which is equal to $\frac{\sigma}{k\epsilon_0}$. The factor of $k$ is because the medium is not vacuum anymore. (Clarification needed because in my book, K does not come in both the expressions for $E_0$ and $E_i$).
Similarly, $E_i=\frac{\sigma'}{k\epsilon_0}$

Hence, $\sigma'=\frac{K-1}{K}\sigma$.

Although I got the answer, in my book, its given $E_0=\frac{\sigma}{\epsilon_0}$ and $E_I=\frac{\sigma'}{k}$. The K term is missing. Is it a printing error or am I wrong?

2. Dec 4, 2015