Learn steps to solve problems of equilibrium in physics. Practice equation of equilibrium of a particle under the action of gravity and electrostatic force & point of equilibrium between two point charges.

- Sum of all external forces acting on a body is zero.

- Gravitational
- Friction
- Spring force
- Tension
- Normal
- Electrostatic

- Sum of all external forces acting on a body is zero.

- Gravitational
- Friction
- Spring force
- Tension
- Normal
- Electrostatic

A \(2\times10^{-3}C\)

B \(3\times10^{-3}C\)

C \(4\times10^{-3}C\)

D \(5\times10^{-3}C\)

- Any charge kept between two point charges of opposite nature cannot be at equilibrium.
- The force exerted due to both the fixed charges act in same direction.

- When both the charges are of same nature.

Under equilibrium

\(F_1 = F_2\)

\(\dfrac {KQq_1}{r_1^2}=\dfrac {KQq_2}{r_2^2}\)

\(\Rightarrow \left ( \dfrac {r_1}{r_2} \right)^2=\dfrac {q_1}{q_2}\)

\(\Rightarrow\dfrac {r_1}{r_2}=\sqrt{\dfrac {q_1}{q_2}}\)

**NOTE: **Distance of zero force between two point charges of same nature are in ratio equal to the square root of ratio of charges.

A \(\dfrac {4}{5}\)

B \(\dfrac {2}{5}\)

C \(\dfrac {3}{5}\)

D \(\dfrac {1}{5}\)

For equilibrium of q_{2}

mg = F_{electrostatic }

_{\(mg=\dfrac {kq_1q_2} {H^2}\)}

_{\(H=\sqrt { \dfrac {kq_1q_2} {mg} }\)}

A \(q=\sqrt {16\pi\epsilon_0 \ell^2\;\text{mg}\; tan\theta.sin^2\theta}\)

B \(q=\sqrt {4\pi\epsilon_0 \ell^2\;\text{mg}\; tan\theta.sin^2\theta}\)

C \(q=\sqrt {4\pi\epsilon_0 \ell\;\text{mg}\; tan\theta.sin^2\theta}\)

D \(q=\sqrt {16\pi\epsilon_0 \ell^2\;\text{mg}\; tan\theta}\)

- Consider two point charges q
_{1}and q_{2}of masses m_{1}and m_{2}respectively attached to a string of same length and suspended from common point.

- External force on both charges are equal. So, the one with greater mass will have less angle at center.

i.e. m_{1} > m_{2}

\(\theta_1<\theta _2\)

**NOTE:** \(\theta\) is dependent only on the mass of charged particle and independent of the magnitude of charge.

A Angle made by \(1\, C\) charge with vertical is more

B Angle made by \(5\,C\) charge with vertical is more

C Both angles are equal

D Can't be determined

When \(|-q_1|=|q_2|\)

- No point of equilibrium.

- When\(|-q_1|>|q_2|\)
- No point of equilibrium in region 2.
- No. equilibrium point in region 1. As on any point in region 1, forces due to q
_{1}and q_{2}will oppose but force exerted by q_{1}is greater as \(|-q_1|\)is greater in magnitude and charge is near to it. - Possible equilibrium in region 3

Similarly,

If \(|-q_1|<|q_2|\)

Equilibrium lies in region 1.

A \(\sqrt{\dfrac {|q_1|}{|q_2|}}\)

B \(\sqrt{\dfrac {|q_2|}{|q_1|}}\)

C \(\sqrt {{|q_1|}\;{|q_2|}}\)

D \(\dfrac {|q_1|}{|q_2|}\)

A \(x=\dfrac {d}{1+\sqrt3}\)

B \(x=\dfrac {d}{1+\sqrt2}\)

C \(\sqrt2\;d\)

D \(\sqrt3\;d\)