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Physics · JEE | Electrostatic Potential and Capacitance

हिन्दी

A linear isotropic dielectric slab is placed in an external uniform electric field \( E_0 \). If the electric susceptibility of the dielectric is \( \chi_e = 4 \), and the magnitude of the polarization vector is \( P = 1.2 \times 10^{-7}\text{ C/m}^2 \), what is the magnitude of the external electric field \( E_0 \) in \( \text{V/m} \)? (Take \( \varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2} \))

  1. A
    \( 1.35 \times 10^4 \)
  2. B
    \( 3.39 \times 10^3 \)
  3. C
    \( 6.78 \times 10^3 \)
  4. D
    \( 2.71 \times 10^4 \)
View Answer & Solution

Correct answer · Option B

\( 3.39 \times 10^3 \)

Explanation

Using the relation \( P = \varepsilon_0 \chi_e E_0 \), we substitute \( P = 1.2 \times 10^{-7} \), \( \chi_e = 4 \), and \( \varepsilon_0 = 8.85 \times 10^{-12} \) to yield \( E_0 = \frac{1.2 \times 10^{-7}}{4 \times 8.85 \times 10^{-12}} \approx 3.39 \times 10^3\text{ V/m} \).
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