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NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEEasy

Question

A parallel plate condenser has a uniform electric field E (V/m) in the space between the plates. If the distance between the plates is d (m) and area of each plate is A (m²), the energy (joule) stored in the condenser is:

A

1/2 ε₀E²

B

ε₀EAd

C

1/2 ε₀E²Ad

D

E²Ad/ε₀

Step-by-Step Solution

The energy stored (UU) in a capacitor is given by the formula U=12CV2U = \frac{1}{2}CV^2. For a parallel plate capacitor, the capacitance is C=ε0AdC = \frac{\varepsilon_0 A}{d} and the potential difference is related to the electric field by V=EdV = Ed.

Substituting these values into the energy equation: U=12(ε0Ad)(Ed)2U = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \right) (Ed)^2 U=12(ε0Ad)(E2d2)U = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \right) (E^2 d^2) U=12ε0E2(Ad)U = \frac{1}{2} \varepsilon_0 E^2 (Ad)

Alternatively, the energy density (energy per unit volume) is u=12ε0E2u = \frac{1}{2}\varepsilon_0 E^2. The total energy is the product of energy density and the volume of the space between the plates (Volume=A×dVolume = A \times d). Thus, U=u×Volume=12ε0E2AdU = u \times Volume = \frac{1}{2}\varepsilon_0 E^2 Ad.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROSTATIC POTENTIAL AND CAPACITANCE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEparallelcondenseruniformelectricbetween

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