Q

# Help me answer: - Electrostatics - JEE Main-10

Two charges, each equal to q, are kept at on the - axis. A particle of mass m and charge is placed at the origin. If charge is given a small displacement - axis,the net force acting on the particle is proportional to :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

85 Views
N

As we discussed in

Magnitude of the Resultant force -

$\dpi{100} F_{net}=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\cos \Theta }$

- wherein

$F_{net} = 2Fcos\theta$

$F_{net} = \frac {2Kq(\frac{q}{2})Y}{(\sqrt{Y^2 + a^2})^2(\sqrt{Y^2 + a^2})}$

$F_{net} = \frac {2Kq(\frac{q}{2})Y}{{Y^2 + a^2})^\frac{3}{2}}=\frac{Kq^2Y}{a^3}$

$\therefore F \propto Y$

Option 1)

Option 2)

Option 3)

Option 4)

Exams
Articles
Questions