back to directory
NEET PHYSICSELECTRIC CHARGES AND FIELDSMedium

Question

Two equal negative charges of charge -q are fixed at the points (0, a) and (0, -a) on the Y-axis. A positive charge Q is released from rest at the point (2a, 0) on the X-axis. The charge Q will:

A

execute simple harmonic motion about the origin.

B

move to the origin and remain at rest.

C

move to infinity.

D

execute oscillatory but not simple harmonic motion.

Step-by-Step Solution

The positive charge QQ is attracted by both fixed negative charges q-q. Due to the symmetry of the setup, the vertical components of the forces cancel out, while the horizontal components add up, creating a net restoring force directed towards the origin.

The magnitude of this net force at a distance xx from the origin is given by the superposition of Coulomb forces: Fnet=2kQqx(x2+a2)3/2F_{net} = \frac{2kQqx}{(x^2 + a^2)^{3/2}}.

For a particle to execute Simple Harmonic Motion (SHM), the restoring force must be directly proportional to the displacement (FxF \propto -x). In this case, the force is proportional to xx only when xax \ll a (small oscillations). However, the charge is released at x=2ax = 2a, which is not small compared to aa. Since the relationship between force and displacement is non-linear for this amplitude, the motion will be oscillatory (repetitive back and forth through the equilibrium point) but not simple harmonic.

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 FIELDSnegativechargeschargepointspositive

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 →