Suppose a uniformly charged wall provides a uniform electric field of N/C normally. A charged particle of mass 2g being suspended through a silk thread of length 20 cm and remains stayed at a distance of 10 cm from the wall. Then the charge on the partic

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Suppose a uniformly charged wall provides a uniform electric field of $2 \times 10^{4}$ N/C normally. A charged particle of mass 2g being suspended through a silk thread of length 20 cm and remains stayed at a distance of 10 cm from the wall. Then the charge on the particle will be $\frac{1}{\sqrt{x}} μ C$ where $x =$ __________. [use g = 10 $m / s^{2}$ ] [2024]

Ans.
(3)

$\sin θ = \frac{10}{20} = \frac{1}{2} \Rightarrow θ = 30 °$

Along x-axis

$\sum {(f_{net})}_{y} = 0 (for equilibrium)$

$T \sin θ = q E$ ...(i)

Along y-axis

$\sum {(f_{net})}_{x} = 0 (for equilibrium)$

$T \cos θ = m g$ ...(ii)

From (i) divide by (ii), $\tan θ = \frac{q E}{m g}$

$q = \frac{m g \tan θ}{E}$

$= \frac{2 \times 10^{- 3} \times 10}{2 \times 10^{4}} \times \frac{1}{\sqrt{3}} = \frac{10^{- 6}}{\sqrt{3}} C$

$\Rightarrow q = \frac{1}{\sqrt{3}} μ C \Rightarrow x = 3$

Want Us to Email you About Special Offers & Updates?