| $$D_{2n} - D_{1n} = \sigma$$ | | $$D_{2n} - D_{1n} = \sigma$$ |
| where $\sigma$ is the free surface charge density at the boundary. | | where $\sigma$ is the free surface charge density at the boundary. |
| Since $D = \varepsilon_0 E$, for the first case ($E_1 = E$, $E_2 = 2E$) we have: | | Since $D = \varepsilon_0 E$, for the first case ($E_1 = E$, $E_2 = 2E$) we have: |
| $$\sigma = \varepsilon_0 E_2 - \varepsilon_0 E_1 = \varepsilon_0 (2E) - \varepsilon_0 E = \varepsilon_0 E$$ | | $$\sigma = \varepsilon_0 E_2 - \varepsilon_0 E_1 = \varepsilon_0 (2E) - \varepsilon_0 E = \varepsilon_0 E$$ |
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| To find the pressure, consider a thin cylindrical shell at the boundary of the two media. Let us isolate a small portion of the surface with charge $\Delta q = \sigma \Delta S$. Let the field generated by this specific portion be $E_0$, and the field generated by all other charges in the system (the external field) be $E_0'$. | | To find the pressure, consider a thin cylindrical shell at the boundary of the two media. Let us isolate a small portion of the surface with charge $\Delta q = \sigma \Delta S$. Let the field generated by this specific portion be $E_0$, and the field generated by all other charges in the system (the external field) be $E_0'$. |
| From the superposition principle, the total fields on either side of the boundary are: | | From the superposition principle, the total fields on either side of the boundary are: |
| $$E_2 = E_0' + E_0 = 2E$$ | | $$E_2 = E_0' + E_0 = 2E$$ |
| $$E_1 = E_0' - E_0 = E$$ | | $$E_1 = E_0' - E_0 = E$$ |
| By adding these two equations, we can find the external field $E_0'$: | | By adding these two equations, we can find the external field $E_0'$: |
| $$2E_0' = 3E \implies E_0' = \frac{3}{2}E$$ | | $$2E_0' = 3E \implies E_0' = \frac{3}{2}E$$ |
| The total force acting on this portion is: | | The total force acting on this portion is: |
| $$\Delta F = \sigma \Delta S E_0'$$ | | $$\Delta F = \sigma \Delta S E_0'$$ |
| Thus, the electrostatic pressure at the interface is: | | Thus, the electrostatic pressure at the interface is: |
| $$P = \frac{\Delta F}{\Delta S} = \sigma E_0' = (\varepsilon_0 E)\left(\frac{3}{2}E\right) = \frac{3\varepsilon_0 E^2}{2}$$ | | $$P = \frac{\Delta F}{\Delta S} = \sigma E_0' = (\varepsilon_0 E)\left(\frac{3}{2}E\right) = \frac{3\varepsilon_0 E^2}{2}$$ |
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| Similarly, for the second case ($E_1 = E$, $E_2 = -2E$), we have: | | Similarly, for the second case ($E_1 = E$, $E_2 = -2E$), we have: |
| $$\sigma = \varepsilon_0 E_2 - \varepsilon_0 E_1 = \varepsilon_0 (-2E) - \varepsilon_0 E = -3\varepsilon_0 E$$ | | $$\sigma = \varepsilon_0 E_2 - \varepsilon_0 E_1 = \varepsilon_0 (-2E) - \varepsilon_0 E = -3\varepsilon_0 E$$ |
| The external field in this case is: | | The external field in this case is: |
| $$E_0' = \frac{E_1 + E_2}{2} = \frac{E - 2E}{2} = -\frac{1}{2}E$$ | | $$E_0' = \frac{E_1 + E_2}{2} = \frac{E - 2E}{2} = -\frac{1}{2}E$$ |
| And the pressure is: | | And the pressure is: |
| $$P = \sigma E_0' = (-3\varepsilon_0 E)\left(-\frac{1}{2}E\right) = \frac{3\varepsilon_0 E^2}{2}$$ | | $$P = \sigma E_0' = (-3\varepsilon_0 E)\left(-\frac{1}{2}E\right) = \frac{3\varepsilon_0 E^2}{2}$$ |
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| <b>Why is the pressure the same?</b> | | <b>Why is the pressure the same?</b> |
| We can obtain a general formula for the pressure $P$ using analogous reasoning. By substituting $E_0' = \frac{E_1 + E_2}{2}$ and $\sigma = \varepsilon_0(E_2 - E_1)$: | | We can obtain a general formula for the pressure $P$ using analogous reasoning. By substituting $E_0' = \frac{E_1 + E_2}{2}$ and $\sigma = \varepsilon_0(E_2 - E_1)$: |
| $$P = \sigma E_0' = \varepsilon_0(E_2 - E_1)\frac{E_1 + E_2}{2} = \frac{\varepsilon_0(E_2^2 - E_1^2)}{2}$$ | | $$P = \sigma E_0' = \varepsilon_0(E_2 - E_1)\frac{E_1 + E_2}{2} = \frac{\varepsilon_0(E_2^2 - E_1^2)}{2}$$ |
| From this final equation, we see that the electrostatic pressure depends on the squares of the electric fields. Therefore, changing the sign of $E_2$ leaves the pressure unchanged. | | From this final equation, we see that the electrostatic pressure depends on the squares of the electric fields. Therefore, changing the sign of $E_2$ leaves the pressure unchanged. |
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| #### Answer | | #### Answer |
| a) $\sigma = \varepsilon_0 E$; $P = \frac{3\varepsilon_0 E^2}{2}$ | | a) $\sigma = \varepsilon_0 E$; $P = \frac{3\varepsilon_0 E^2}{2}$ |
| | | |
| b) $\sigma = -3\varepsilon_0 E$; $P = \frac{3\varepsilon_0 E^2}{2}$ | | b) $\sigma = -3\varepsilon_0 E$; $P = \frac{3\varepsilon_0 E^2}{2}$ |
