Question:
In a parallel plate capacitor, the plate at \( x = 0 \) is grounded and the plate at \( x = d \) is maintained at a potential \( V_0 \). The space between the two plates is filled with a linear dielectric of permittivity \( \epsilon = \epsilon_0 \left(1 + \frac{x}{d}\right) \), where \( \epsilon_0 \) is the permittivity of free space. Neglecting the edge effects, the electric field (\(\vec{E}\)) inside the capacitor is .
In a parallel plate capacitor, the plate at \( x = 0 \) is grounded and the plate at \( x = d \) is maintained at a potential \( V_0 \). The space between the two plates is filled with a linear dielectric of permittivity \( \epsilon = \epsilon_0 \left(1 + \frac{x}{d}\right) \), where \( \epsilon_0 \) is the permittivity of free space. Neglecting the edge effects, the electric field (\(\vec{E}\)) inside the capacitor is .
Updated On: Jul 12, 2024
- \(-\frac{V_0}{(d+x)ln 2}\hat{x}\)
- \(-\frac{V_0}{d}\hat{x}\)
- \(-\frac{V_0}{(d+x)}\hat{x}\)
- \(-\frac{V_0d}{(d+x)x}\hat{x}\)
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The Correct Option is A
Solution and Explanation
The correct option is (A) :\(-\frac{V_0}{(d+x)ln 2}\hat{x}\)
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