Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density +sigma$+\sigma$ and -2sigma$-2\sigma$. The force experienced by a point charge +q$+q$ placed at the mid point between two plates will be:

### Related Formula E = fracsigma_textinnerepsilon_0$$E = \frac{\sigma_{\text{inner}}}{\epsilon_0}$$ ### Core Logic For parallel conducting plates of large area, charges redistribute on the outer and inner faces to maintain electrostatic equilibrium. Total charge per unit area on Plate 1: q_1 = sigma$q_1 = \sigma$ Total charge per unit area on Plate 2: q_2 = -2sigma$q_2 = -2\sigma$ The outer surface charge density on the far sides of both plates must be equal: sigma_textouter = fracq_1 + q_22 = fracsigma - 2sigma2 = -fracsigma2$$\sigma_{\text{outer}} = \frac{q_1 + q_2}{2} = \frac{\sigma - 2\sigma}{2} = -\frac{\sigma}{2}$$ Now, compute the charges on the inner facing surfaces: - Inner face of Plate 1: sigma_textinner1 = q_1 - sigma_textouter = sigma - left(-fracsigma2right) = frac3sigma2$$\sigma_{\text{inner1}} = q_1 - \sigma_{\text{outer}} = \sigma - \left(-\frac{\sigma}{2}\right) = \frac{3\sigma}{2}$$ - Inner face of Plate 2: sigma_textinner2 = q_2 - sigma_textouter = -2sigma - left(-fracsigma2right) = -frac3sigma2$$\sigma_{\text{inner2}} = q_2 - \sigma_{\text{outer}} = -2\sigma - \left(-\frac{\sigma}{2}\right) = -\frac{3\sigma}{2}$$ In the region between the plates, both inner surfaces create an electric field in the same direction (away from the positive plate 1 and towards negative plate 2): E = fracsigma_textinner12epsilon_0 + frac|sigma_textinner2|2epsilon_0 = frac3sigma/22epsilon_0 + frac3sigma/22epsilon_0 = frac3sigma2epsilon_0$$E = \frac{\sigma_{\text{inner1}}}{2\epsilon_0} + \frac{|\sigma_{\text{inner2}}|}{2\epsilon_0} = \frac{3\sigma/2}{2\epsilon_0} + \frac{3\sigma/2}{2\epsilon_0} = \frac{3\sigma}{2\epsilon_0}$$ Thus, the electrostatic force on +q$+q$ is: F = q E = frac3sigma q2epsilon_0$$F = q E = \frac{3\sigma q}{2\epsilon_0}$$ ### Step 1: Final Conclusion The force experienced by the point charge +q$+q$ is frac3sigma q2epsilon_0$\frac{3\sigma q}{2\epsilon_0}$. ### Pattern Recognition For conducting plates with total charges Q_1$Q_1$ and Q_2$Q_2$, always calculate the outer charge first: Q_textouter = fracQ_1+Q_22$Q_{\text{outer}} = \frac{Q_1+Q_2}{2}$. The field inside the gap is exclusively due to the inner surfaces: E = fracsigma_textinnerepsilon_0$E = \frac{\sigma_{\text{inner}}}{\epsilon_0}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics