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

Illustration Questions

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.

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

Hence, option B is incorrect.

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

Hence, option C is incorrect.

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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.

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

A image
B image
C image
D image

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

Illustration Questions

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

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On x-axis

F = Fe

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

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

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

\(q^2=4\times10^{-6}\)

\(q=2\times10^{-3}C\)

 

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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].

image
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. 

Illustration Questions

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

\(F_1 = F_2\)

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

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

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

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\(\dfrac {r_1}{r_2}=\sqrt{\dfrac {4}{25}}\)

\(\dfrac {r_1}{r_2}=\dfrac {2}{5}\)

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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

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

For equilibrium of q2

mg = Felectrostatic 

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

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

Illustration Questions

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

image

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

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

where, \(m\) = mass of  \(q_1\)

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

\(H\) = distance between two charges

 

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

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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.

image
A

10 m

.

B

20 m

C

30 m

D

40 m

Option C is Correct

Illustration Questions

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 \(\theta\)

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On Y-axis

mg = T sin (90 – \(\theta\))

mg = T cos \(\theta\)

image

From figure,

 \(tan\theta = \dfrac {F}{mg}\)

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

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Solve for q

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

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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.

image
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

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.

Illustration Questions

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, \(\theta_1>\theta_2\)

 

 

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

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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

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.

Illustration Questions

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.

image
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

Illustration Questions

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.

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

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

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From (i) and (ii)

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

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

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

image
A

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

.

B

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

C

\(\sqrt2\;d\)

D

\(\sqrt3\;d\)

Option B is Correct

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