Informative line

### Charges Under Equilibrium

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.

# Condition of Equilibrium

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

## Types of Forces acting on a Particle

1. Gravitational
2. Friction
3. Spring force
4. Tension
5. Normal
6. Electrostatic

## Steps to solve Problems of Equilibrium

#### In which of the following situation, the given particle 1 is in equilibrium?

A

B

C

D

×

Net force = $$F_1 –F$$

$$=\dfrac {kq_1q_2}{r^2}-mg$$

$$=\dfrac {9\times10^9\times1\times10^{-6}\times5} {(30)^2}-5\times10$$

=$$0$$ (Equilibrium)

Hence, option A is correct.

Both F12 and F13 are in same direction. So, net force will not be zero.

Hence, option B is incorrect.

Since net force in x-direction is not zero. So, total net force is not zero.

Hence, option C is incorrect.

Forces in y-direction i.e. F13 and F12 nullify each other. But forces in x-direction will not be zero. So, net force is not zero.

Hence, option D is incorrect.

### In which of the following situation, the given particle 1 is in equilibrium?

A
B
C
D

Option A is Correct

# Calculation of Charge on Blocks such that they remain at Rest

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

## Types of forces acting on a particle

1. Gravitational
2. Friction
3. Spring force
4. Tension
5. Normal
6. Electrostatic

## Steps to solve problems of equilibrium

#### Two blocks A and B of mass m = 2 kg each are kept d = 30 m apart on a horizontal rough surface. Calculate the maximum charge 'q' such that both the blocks remain at rest [ $$\mu$$= 0.2].

A $$2\times10^{-3}C$$

B $$3\times10^{-3}C$$

C $$4\times10^{-3}C$$

D $$5\times10^{-3}C$$

×

On y-axis

N = mg

On x-axis

F = Fe

$$\Rightarrow\mu N=\dfrac {kq_1q_2}{r^2}$$

$$\Rightarrow \mu \text{ mg}=\dfrac {kq_1q_2}{r^2}$$

$$\Rightarrow 0.2\times20\times10=\dfrac {9\times10^9\times q^2}{(30)^2}$$

$$q^2=4\times10^{-6}$$

$$q=2\times10^{-3}C$$

### Two blocks A and B of mass m = 2 kg each are kept d = 30 m apart on a horizontal rough surface. Calculate the maximum charge 'q' such that both the blocks remain at rest [ $$\mu$$= 0.2].

A

$$2\times10^{-3}C$$

.

B

$$3\times10^{-3}C$$

C

$$4\times10^{-3}C$$

D

$$5\times10^{-3}C$$

Option A is Correct

# Point of equilibrium between two point charge

### Case-I

• 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.

### Case-II

• 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.

#### Two point charges, q1 = 4 C and q2 = 25 C are separated by some distance. If there is any point of zero force between them, then calculate the ratio of distance of zero point from q1 and q2?

A $$\dfrac {4}{5}$$

B $$\dfrac {2}{5}$$

C $$\dfrac {3}{5}$$

D $$\dfrac {1}{5}$$

×

When both the charges are of same nature

Under equilibrium

$$F_1 = F_2$$

$$\dfrac {KQq_1}{r_1^2}=\dfrac {KQq_2}{r_2^2}$$

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

$$\Rightarrow\dfrac {r_1}{r_2}=\sqrt{\dfrac {q_1}{q_2}}$$

$$\dfrac {r_1}{r_2}=\sqrt{\dfrac {4}{25}}$$

$$\dfrac {r_1}{r_2}=\dfrac {2}{5}$$

### Two point charges, q1 = 4 C and q2 = 25 C are separated by some distance. If there is any point of zero force between them, then calculate the ratio of distance of zero point from q1 and q2?

A

$$\dfrac {4}{5}$$

.

B

$$\dfrac {2}{5}$$

C

$$\dfrac {3}{5}$$

D

$$\dfrac {1}{5}$$

Option B is Correct

# Comparison of Angle made by Two Charges Hanging with String

• Consider two point charges q1 and q2 of masses m1 and m2 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. m1 > m2

$$\theta_1<\theta _2$$

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

#### Consider a charged particle of mass m1 = 2 kg and charge q1 = 5C and another particle of mass m2 =  5 kg and charge q2 = 1 C are suspended to a common point with two strings of equal length. Choose the correct statement.

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

×

Since, m2 > m1

So, $$\theta_1>\theta_2$$

So, angle made by 5C charge with vertical is more.

### Consider a charged particle of mass m1 = 2 kg and charge q1 = 5C and another particle of mass m2 =  5 kg and charge q2 = 1 C are suspended to a common point with two strings of equal length. Choose the correct statement.

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

Option B is Correct

# Equilibrium of a Particle under the Action of Gravity and Electrostatic Force

For equilibrium of q2

mg = Felectrostatic

$$mg=\dfrac {kq_1q_2} {H^2}$$

$$H=\sqrt { \dfrac {kq_1q_2} {mg} }$$

#### Consider two charged particles q1 and q2 as shown in figure. Calculate H for the equilibrium of q2 .  Given mass m = 5 kg, q1 = 5C and q2 = 1$$\mu$$C.

A 10 m

B 20 m

C 30 m

D 40 m

×

Under equilibrium

mg = Felectrostatic

$$mg=\dfrac {kq_1q_2} {H^2}$$

$$H= \sqrt{\dfrac {kq_1q_2} {mg}}$$

where, $$m$$ = mass of  $$q_1$$

$$k=\dfrac {1}{4\pi\epsilon_0}=\text {constant}$$

$$H$$ = distance between two charges

$$H=\sqrt{\dfrac {9\times10^9\times1\times10^{-6}\times5}{5\times10}}=30\;\text{m}$$

### Consider two charged particles q1 and q2 as shown in figure. Calculate H for the equilibrium of q2 .  Given mass m = 5 kg, q1 = 5C and q2 = 1$$\mu$$C.

A

10 m

.

B

20 m

C

30 m

D

40 m

Option C is Correct

#### Two identical charged spheres each of mass m are hanging in equilibrium. Length of each string is  $$\ell$$ and angle between strings is 2$$\theta$$. Find the magnitude of charge on each sphere.

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

×

On X-axis

F = T cos (90 – $$\theta$$)

F = T sin $$\theta$$

On Y-axis

mg = T sin (90 – $$\theta$$)

mg = T cos $$\theta$$

From figure,

$$tan\theta = \dfrac {F}{mg}$$

$$tan\theta=\dfrac {kq_1q_2} {(2\ell\; sin\theta)^2\;\text{mg}}$$

Solve for q

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

### Two identical charged spheres each of mass m are hanging in equilibrium. Length of each string is  $$\ell$$ and angle between strings is 2$$\theta$$. Find the magnitude of charge on each sphere.

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

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

Option A is Correct

# Point of Equilibrium on Line Joining Two Point Charges of Opposite Sign

### Case-I

When $$|-q_1|=|q_2|$$

• No point of equilibrium.

### Case-II

• 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 q1 and q2 will oppose but force exerted by q1 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.

#### Two point charges -q1 and q2,  are separated by some distance. Calculate the ratio of distance of equilibrium point from charges.

A $$\sqrt{\dfrac {|q_1|}{|q_2|}}$$

B $$\sqrt{\dfrac {|q_2|}{|q_1|}}$$

C $$\sqrt {{|q_1|}\;{|q_2|}}$$

D $$\dfrac {|q_1|}{|q_2|}$$

×

For equilibrium of Q

F1 = F2

$$\dfrac {k|-q_1|Q}{r_1^2}=\dfrac {k\;|q_2|\;Q}{r_2^2}$$

$$\dfrac {r_1}{r_2}=\sqrt{\dfrac {|q_1|}{|q_2|}}$$

### Two point charges -q1 and q2,  are separated by some distance. Calculate the ratio of distance of equilibrium point from charges.

A

$$\sqrt{\dfrac {|q_1|}{|q_2|}}$$

.

B

$$\sqrt{\dfrac {|q_2|}{|q_1|}}$$

C

$$\sqrt {{|q_1|}\;{|q_2|}}$$

D

$$\dfrac {|q_1|}{|q_2|}$$

Option A is Correct

#### Calculate the position of third charge in between the two given charges so that all the three charges are in equilibrium.

A $$x=\dfrac {d}{1+\sqrt3}$$

B $$x=\dfrac {d}{1+\sqrt2}$$

C $$\sqrt2\;d$$

D $$\sqrt3\;d$$

×

For Q to lie between q and 2q, it should be negative.

Then, for net force on q to be zero; $$\dfrac {kqQ}{x^2}=\dfrac {kq(2q)}{d^2}$$             ....(i)

For net force on 2q to be zero; $$\dfrac {kQ2q}{(d-x)^2}=\dfrac {kq(2q)}{d^2}$$  ...(ii)

From (i) and (ii)

$$\dfrac {1}{x^2}=\dfrac {2}{(d-x)^2}$$

$$\Rightarrow x=\dfrac {d}{1+\sqrt 2}$$

### Calculate the position of third charge in between the two given charges so that all the three charges are in equilibrium.

A

$$x=\dfrac {d}{1+\sqrt3}$$

.

B

$$x=\dfrac {d}{1+\sqrt2}$$

C

$$\sqrt2\;d$$

D

$$\sqrt3\;d$$

Option B is Correct