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NEET PHYSICSEasy

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0×10−22 C/m217.0 \times 10^{-22} \, \text{C/m}^2. The electric field between the plates is:

A

0.96×10−10 N/C0.96 \times 10^{-10} \, \text{N/C}

B

1.92×10−10 N/C1.92 \times 10^{-10} \, \text{N/C}

C

0

D

3.84×10−10 N/C3.84 \times 10^{-10} \, \text{N/C}

Step-by-Step Solution

For two large, parallel, thin metal plates with surface charge densities of opposite signs (σ\sigma and −σ-\sigma), the electric field in the outer regions is zero. In the inner region between the plates, the electric fields due to both plates add up because they point in the same direction (away from positive, towards negative).

The magnitude of the field is given by E=σ2ε0+σ2ε0=σε0E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}.

Given: σ=17.0×10−22 C/m2\sigma = 17.0 \times 10^{-22} \, \text{C/m}^2 ε0=8.854×10−12 C2N−1m−2\varepsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2\text{N}^{-1}\text{m}^{-2}

Calculation: E=17.0×10−228.854×10−12E = \frac{17.0 \times 10^{-22}}{8.854 \times 10^{-12}} E≈1.92×10−10 N/CE \approx 1.92 \times 10^{-10} \, \text{N/C}. (See NCERT Physics Class 12, Exercise 1.24).

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