NVAMediumJEE 2023Capacitors & Dielectrics

JEE Physics 2023 Question with Solution

A parallel plate capacitor with plate area AA and plate separation dd is filled with a dielectric material of dielectric constant K=4K = 4. The thickness of the dielectric material is xx, where x<dx < d. Let C1C_1 and C2C_2 be the capacitance of the system for x=13dx = \frac{1}{3}d and x=23dx = \frac{2}{3}d, respectively. If C1=2μFC_1 = 2 \, \mu\text{F}, the value of C2C_2 is in μF\mu\text{F}:

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: A parallel plate capacitor has dielectric constant K=4K = 4. For x=13dx = \frac{1}{3}d, the capacitance is C1=2μFC_1 = 2 \, \mu\text{F}.

Find: The value of C2C_2 for x=23dx = \frac{2}{3}d.

From the solution, the stated result is that the value of C2C_2 is 3μF3 \, \mu\text{F}.

Therefore, the required numerical answer is 33.

Common mistakes

  • Treating the partially filled capacitor as if the entire separation dd is filled with dielectric is incorrect, because only thickness xx contains dielectric. Use the effective series combination of dielectric-filled and air-filled regions instead.

  • Substituting x=13x = \frac{1}{3} or x=23x = \frac{2}{3} directly instead of x=13dx = \frac{1}{3}d and x=23dx = \frac{2}{3}d is wrong. The thickness must remain in the same dimensional form as the plate separation dd.

  • Writing the final NVA answer with units, such as 3μF3 \, \mu\text{F}, is not accepted in the answer field. The answer field must contain only the number 33.

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