Question

Suppose, we have a parallel plate capacitor and we know the following about it: Area of each plate = $0.0012m^2$ Distance between the two plates = $0.002m$ Charge on each plate after fully charging the capacitor = $2\times10^{-6}C$ Potential difference between the plates after fully charging the capacitor = $4\times 10^{-3}V$ My solution: We know that the electric field intensity at a point between two equally and oppositely charged plates is $\LARGE\frac{\sigma}{\epsilon_0}$. So, for the above case, $$\large E = \frac{\sigma}{\epsilon_0}$$ $$\large\implies E = \frac{\frac{2\times10^{-6}}{0.0012}}{8.85\times10^{-12}}{NC^{-1}}$$ $$\large\implies E =188323917.1NC^{-1}$$ My book's solution: $$E = \frac{V}{d}$$ $$\implies E=\frac{4\times 10^{-3}V}{0.002m}NC^{-1}$$ $$\implies E = 2NC^{-1}$$ If the books solution is correct, could you please explain why my answer is wrong and the book's is correct? Thanks in advance!

Answer 1

Answer:

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