Verified Solution • 2025–26 Session moderate

QJEEM18Ph0c34: JEE Main 2018 Physics - Class 12 JEEM18Ph0c34 Solution: A body of mass M and charge q is connected to spring of spring…

A body of mass

M

and charge

q

is connected to spring of spring constant

k .

It is oscillating along

x

-direction about its equilibrium position, taken to be at

x = 0,

with an amplitude

A

. An electric field

E

is applied along the

x

-direction. Which of the following statements is correct ?

Multiple Choice Options

A The new equilibrium position is at a distance qE2k from x=0.
B The total energy of the system is 12mω2A2+12q2E2k.
C The total energy of the system is 12mω2A2−12q2E2k.
D The new equilibrium position is at a distance 2qEk from x=0.

Step-by-Step Verified Solution

On the body of charge q a electric fied E is applied, because of this equilibrium position of body will shift to a point where resulttant force is zero.

∴

kx_eq = qE

\Rightarrow

x_eq =

\frac{q E}{K}

∴

Total energy of the system

=

\frac{1}{2} m ω^{2} A^{2} + \frac{1}{2} K x_{e q}^{2}

\frac{1}{2}

ω^{2}

A² +

\frac{1}{2} . \frac{q^{2} E^{2}}{K}

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