back to directory
NEET PHYSICSELECTRIC CHARGES AND FIELDSEasy

Question

Two infinitely long parallel conducting plates having surface charge densities +\sigma and −\sigma respectively, are separated by a small distance. The medium between the plates is a vacuum. If ε₀ is the dielectric permittivity of vacuum, then the electric field in the region between the plates is:

A

0 V/m

B

\sigma /(2ε₀) V/m

C

\sigma /ε₀ V/m

D

2\sigma /ε₀ V/m

Step-by-Step Solution

The electric field due to a single uniformly charged infinite plane sheet is E=σ2ϵ0E = \frac{\sigma}{2\epsilon_0} . For two parallel plates with surface charge densities +σ+\sigma and σ-\sigma:

  1. In the region between the plates, the electric field due to the positive plate points away from it (towards the negative plate), and the field due to the negative plate points towards it (also away from the positive plate).
  2. Since both fields point in the same direction, they add up: Enet=E++E=σ2ϵ0+σ2ϵ0=σϵ0E_{net} = E_+ + E_- = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} . (Note: In the outer regions, the fields are opposite and cancel out, resulting in zero electric field).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTRIC CHARGES AND FIELDS. 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.

PHYSICSELECTRIC CHARGES AND FIELDSinfinitelyparallelconductingplateshaving

More ELECTRIC CHARGES AND FIELDS Questions

View all

Two identical charged spheres suspended from a common point by two massless strings of lengths l are initially at a distance d (d << l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as:

A.v ∝ x
B.v ∝ x⁻¹/²
C.v ∝ x⁻¹
D.v ∝ x¹/²
HardSolve

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 $17.0 \times 10^{-22} \, \text{C/m}^2$. The electric field between the plates is:

A.$0.96 \times 10^{-10} \, \text{N/C}$
B.$1.92 \times 10^{-10} \, \text{N/C}$
C.0
D.$3.84 \times 10^{-10} \, \text{N/C}$
EasySolve

Two point charges $q_A = 3 \, \mu\text{C}$ and $q_B = -3 \, \mu\text{C}$ are located $20 \, \text{cm}$ apart in a vacuum. The electric field at the midpoint $O$ of the line $AB$ joining the two charges is:

A.$4.5 \times 10^6 \, \text{N/C}$ along $OA$
B.$5.4 \times 10^6 \, \text{N/C}$ along $OA$
C.$4.5 \times 10^6 \, \text{N/C}$ along $OB$
D.$5.4 \times 10^6 \, \text{N/C}$ along $OB$
EasySolve

The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius 'a' centered at the origin of the field will be given by:

A.4πε₀Aa²
B.πε₀Aa²
C.4πε₀Aa³
D.ε₀Aa²
MediumSolve

The electric field due to a uniformly charged solid sphere of radius R as a function of the distance from its centre is represented graphically by -

A.Option 1
B.Option 2
C.Option 3
D.Option 4
MediumSolve

Four-point +ve charges of the same magnitude (Q) are placed at four corners of a rigid square frame as shown in the figure. The plane of the frame is perpendicular to Z-axis. If a –ve point charge is placed at a distance z away from the above frame (z<<L) then

A.– ve charge oscillates along the Z-axis.
B.It moves away from the frame.
C.It moves slowly towards the frame and stays in the plane of the frame.
D.It passes through the frame only once.
MediumSolve

Shown below is a distribution of charges. The flux of electric field due to these charges through the surface S is:

A.3q/ε₀
B.2q/ε₀
C.q/ε₀
D.Zero
EasySolve

Two point charges A and B, having charges +Q and −Q respectively, are placed at a certain distance apart and the force acting between them is F. If 25% charge of A is transferred to B, then the force between the charges becomes:

A.4F/3
B.F
C.9F/16
D.16F/9
MediumSolve

This neet physics practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice physics sets →