Two charges, each equal to q, are kept at x = - a and x = a on the x-axis. A particle of mass m and charge q0=(q/2) is placed at the origin. If charge q0 is given a small displacement y ( y

Question

Two charges, each equal to q, are kept at x = - a and x = a on the x-axis. A particle of mass m and charge q0=(q/2) is placed at the origin. If charge q0 is given a small displacement y ( y << a) along the y-axis, the net force acting on the particle is proportional to (A) y (B) -y (C) (1/y) (D) -(1/y)

Tardigrade Admin · Accepted Answer

Fnet =2F cos θ Fnet =(2kq((q/2))/(√y2+a2)2).(y/√y2+a2) Fnet =(2kq((q/2))y/(√y2+a2)3/2) ⇒ (kq2y/a3) propto y

$y$

$- y$

$\frac{1}{y}$

$- \frac{1}{y}$

$F_{n e t} = 2 F cos θ$
$F_{n e t} = \frac{2 k q ( \frac{q}{2} )}{( y ^{2} + a ^{2} ) ^{2}} . \frac{y}{y ^{2} + a ^{2}}$
$F_{n e t} = \frac{2 k q ( \frac{q}{2} ) y}{( y ^{2} + a ^{2} ) ^{3/2}}$
$\Rightarrow \frac{k q ^{2} y}{a ^{3}} \propto y$

Q. Two charges, each equal to q, are kept at x=−a and x=a on the x−axis. A particle of mass m and charge q0​=2q​ is placed at the origin. If charge q0​ is given a small displacement y(y<<a) along the y-axis, the net force acting on the particle is proportional to

y

−y

y1​

−y1​

Fnet​=2Fcosθ Fnet​=(y2+a2​)22kq(2q​)​.y2+a2​y​ Fnet​=(y2+a2​)3/22kq(2q​)y​ ⇒a3kq2y​∝y

$y$

$- y$

$\frac{1}{y}$

$- \frac{1}{y}$

$F_{n e t} = 2 F cos θ$
$F_{n e t} = \frac{2 k q ( \frac{q}{2} )}{( y ^{2} + a ^{2} ) ^{2}} . \frac{y}{y ^{2} + a ^{2}}$
$F_{n e t} = \frac{2 k q ( \frac{q}{2} ) y}{( y ^{2} + a ^{2} ) ^{3/2}}$
$\Rightarrow \frac{k q ^{2} y}{a ^{3}} \propto y$