Coulomb’s Law MCQ Quiz - Objective Question with Answer for Coulomb’s Law - Download Free PDF

Calculation:
Let the charge on each sphere A and B be q and the separation be d.

Therefore, the force between spheres A and B is:

F = (1 / (4πɛ₀)) × (q² / d²) ... (1)

When spheres A and C are touched and then separated, charge on each will be:

(q + 0) / 2 = q / 2

Now sphere B is touched with sphere C. Charge on each will be:

(q + q/2) / 2 = (3q) / 4

Now the force between sphere A and sphere B will be:

F' = (1 / (4πɛ₀)) × (q/2 × 3q/4) / d²

= (3/8) × (1 / (4πɛ₀)) × (q² / d²)

⇒ F' = (3/8) × F

Calculation:

\(\phi=-2 \times 10^{4} \frac{\mathrm{Nm}^{2}}{\mathrm{C}}\)

r = 8.0 cm

\(\phi=\frac{\mathrm{q}}{\epsilon_{0}} \Rightarrow \mathrm{q}=\epsilon_{0} \phi \)

= (8.85 × 10^–12) × (–2 × 10⁴)

q = –17.7 × 10^–8 C

The correct answer is - the force between the charges becomes one-fourth and the potential energy becomes half.

Key Points

• Force between charges
  • The force between two point charges is given by Coulomb's law: F = k ×  |q1 × q2| / r^2, where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.
  • When the distance r is doubled, the new force F' becomes: F' = k × |q1 × q2| / (2r)^2 = F / 4.
  • Thus, the force between the charges becomes one-fourth.
• Potential Energy
  • The electric potential energy U between two point charges is given by: U = k × |q1 × q2| / r.
  • When the distance r is doubled, the new potential energy U' becomes: U' = k × |q1 × q2| / (2r) = U / 2.
  • Therefore, the potential energy becomes one-half.

Additional Information

• Coulomb's Law
  • This law states that the force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.
  • It is mathematically expressed as: F = k × |q1 × q2| / r^2.
  • The constant of proportionality k is approximately equal to 8.99 × 10^9 N m^2 C^-2.
• Potential Energy in Electrostatics
  • The potential energy of a system of point charges is the work done to assemble the charges from infinity to their given positions.
  • For two point charges, it is given by: U = k × |q1 × q2| / r.
  • This concept is crucial in understanding the behavior of charges in electrostatic fields.

Calculation:

\(\mathrm{F}=\frac{\mathrm{k}\left(\frac{\theta}{2}\right)\left(\frac{\theta}{2}\right)}{\mathrm{r}^{2}}\)

\(9 \times 10^{-3}=\frac{9 \times 10^{9} \times\left(4 \times 10^{-8}\right) \times 4 \times 10^{-8}}{4 \times \mathrm{r}^{2}}\)

\(\mathrm{r}^{2}=\frac{9 \times 10^{9} \times 16 \times 10^{-16}}{4 \times 9 \times 10^{-3}}=4 \times 10^{-4}\)

r = 2 × 10^–2 m ⇒ 2 cm

Concept:

The electrostatic force (F) between two point charges q₁ and q₂ separated by a distance r in vacuum is given by Coulomb’s law:

F = k (q₁ q₂) / r²

Calculation:

Initially, each charge is q and they are placed at a distance d apart. The force between them is:

F = k (q × q) / d² = k q² / d²

Now, each charge is doubled, so each becomes 2q. To keep the force unchanged (still F), let the new distance be r. The new force F′ is:

F′ = k ((2q) × (2q)) / r² = k (4q²) / r² = 4k q² / r²

We want F′ = F, so:

4k q² / r² = k q² / d²

4 / r² = 1 / d²

r² = 4d² 

⇒ r = 2d

∴ The distance should be changed to 2d to keep the force the same.

CONCEPT:

Coulomb's law in Electrostatics –

• Coulomb's law state’s that force of interaction between two stationary point charges is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them and acts along the straight line joining the two charges.

Force (F) ∝ q₁ × q₂

\(F \propto \;\frac{1}{{{r^2}}}\)

\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)

Where K is a constant = 9 × 10⁹ Nm²/C²

EXPLANATION:

Given – q₁ = 2 × 10^-7 C, q₂ = 3 × 10^-7 C and r = 30 cm = 30 × 10^-2 m

Force is equal to

\(F = \left( {9{\rm{\;}} \times {\rm{\;}}{{10}^9}} \right)\times \frac{{2 \times {{10}^{ - 7}} \times 3 \times {{10}^{ - 7}}}}{{{{\left( {30 \times {{10}^{ - 2}}} \right)}^2}}}\)

\( \Rightarrow F = \frac{{54 \times {{10}^{ - 5}}}}{{900 \times {{10}^{ - 4}}}} = 6 \times {10^{ - 3}}N\)

Concept:

• Coulomb law: If two charges are placed at points separated by distance r force between charges is

1. Directly proportional to the product of charges.
2. Inversely proportional to the square of the distance

\(F=\frac{1}{4\piϵ_0}\frac{q_!q_2}{r^2}\)

• The dielectric constant is a ratio of the permittivity of material to the permittivity of vacuum.

\(K=\frac{ϵ}{ϵ_0}\)  Where,ϵ - permittivity of material and ϵ₀ - permittivity of vacuum. 

Explanation:

• In the first condition, charges are in the air

\(F=\frac{1}{4\piϵ_0}\frac{q_!q_2}{r^2} \) .................. (1)

• Then charges are placed in a material where force is 4F

​\(4F=\frac{1}{4\piϵ}\frac{q_!q_2}{r_1^2}\).....................(2)

• The dielectric constant(K) is 16 therefore permittivity of a material is given by
• ϵ = K ϵ₀ = 16 ϵ₀ 
• Substitute value of ϵ ​in equation 2

\(4F=\frac{1}{4\pi16ϵ_0}\frac{q_!q_2}{r_1^2}\) ..............(3)

• Divide equation 1 by equation 3

\(\frac{1}{4}= \frac{16r_1^2}{r^2}\)

​\(\frac{r_1^2}{r^2}=\frac{1}{64}\)

\(r_1=\frac{r}{8}\)

• Therefore option 1 is correct.

CONCEPT:

• Coulomb’s law: When two charged particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.
  
  Force (F) ∝ q1 × q₂
  \(F \propto \frac{1}{{{r^2}}}\)
  \(F = K\frac{{{q_1} × {q_2}}}{{{r^2}}}\)
  Where K is a constant = 9 × 109 Nm2/C2

CALCULATION:

Given that:

Two α-particles are there, so both charges are equal: q1 = q2 = q

Charge on one alpha particles = q = + (2 × charge on an electron) = +2e 

q = q₁ = q₂ = 2 × 1.6 × 10^-19 C

Distance between them (r) = 3.2 × 10-15 m

\(F = K\frac{{{q^2}}}{{{r^{2\;}}}} = \frac{9\times 10^9 \times (2 \times 1.6 \times 10^{-19})^2}{(3.2\times 10^{-15})^2}\)

F = 90 N

• The Force between two alpha particles will be equal to 90 N.
• Hence option 4 is correct.

Coulomb's law in Electrostatics:

It state’s that force of interaction between two stationary point charges is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them and acts along the straight line joining the two charges.

F ∝ q1 × q2

\(F \propto \;\frac{1}{{{r^2}}}\)

\(F = K\frac{{{q_1}\; \times \;{q_2}}}{{{r^2}}}\)

Where \(K = \frac{1}{4\pi \epsilon_o}\)constant called electrostatic force constant.or Coulomb's Constant

• The value of K depends on the nature of the medium between the two charges and the system of units chosen
• From above it is clear that the constant in Coulomb's Law equation is \( \frac{1}{4\pi \epsilon_o}=9\times 10^ 9 Nm^2/C^2.\) , where εo is the permittivity of free space. 

Concept:

Coulomb’s law:

• When two charged particles of charges q₁ and q₂ are separated by a distance r from each other then the electrostatic force between them is directly proportional to multiplication of charges of two particles and inversely proportional to square of distance between them.

Force (F) ∝ q₁ × q₂

\(F\; \propto \;\frac{1}{{{r^2}}}\)

\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)

Where K is a constant = 9 × 10⁹ Nm²/C²

Explanation:

Given that:

Charge, q₁ = 12 C and q₂ = - 6 C

Distance = r = 3m

Force = F = \(K\frac{{{q_1} \times {q_2}}}{{{r^2}}} = \;9 \times {10^9}\frac{{12 \times \left( { - 6} \right)}}{{{3^2}}}\)

Force = F = -72 × 10⁹ N

The negative sign shows the attractive nature of the force.

CONCEPT:

• Coulumb’s law in Electrostatics: It state’s that force of interaction between two stationary point charges is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them and acts along the straight line joining the two charges.

F ∝ q₁ × q₂

\(F \propto \frac{1}{{{r^2}}}\)

\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)

Where K is a constant = 9 × 10⁹ Nm²/C²

Explanation:

Given – Charge q₁ = q₂ = q, r₁ = r/3

The force between the two charges is,

\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}} = \frac{{K\;{q^2}}}{{{r^2}}}\)

The force between the two charges when the distance is reduced,

\({F_1} = K\frac{{{q_1} \times {q_2}}}{{r_1^2}} = K\frac{{q \times q}}{{{{\left( {\frac{r}{3}} \right)}^2}}} = \frac{{9K\;{q^2}}}{{{r^2}}} = 9F{\rm{\;}}\left[ {F = \frac{{K\;{q^2}}}{{{r^2}}}} \right]\)

The correct answer is option 1) i.e. 2μC.

CONCEPT:

• Electric charge is a property of particles by which they have a tendency to attract or repel each other without touching. 
  • Electric force: Particles possessing opposite charges attract each other and particles possessing like charges repel each other. This force of attraction or repulsion is called electric force.
• Coulomb's law: The relationship between the electric charges and electric force is given by Coulomb's law.
  • Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the distance between the two objects.

The electric force is given as:

 \(F = \frac{kq_1 q_2}{r^2}\)

​Where F is the electric force acting between the two charges, q₁ and q₂ are the two charges, r is the center to center distance between the two objects and k is the proportionality constant known as the Coulomb's law constant and is equal to 9 \(\times \)10⁹ Nm²/C².

CALCULATION:

Given that:

The two charges q₁ and q₂ are identical \(\Rightarrow q_1=q_2=q\)

Electric force (F) of repulsion between the charges = 10 g wt = 10 \(\times \)10^-3 kg \(\times \)10 m/s² = 0.1 N

Distance between the two charges (r) = 0.6 m

\(F = \frac{kq_1 q_2}{r^2}\) \(=\frac{kq^2}{r^2}\)

\(0.1 = \frac{(9 \times 10^9) q^2}{0.6^2}\)

\(q = \sqrt{\frac{0.1 \times 0.6^2}{9\times10^9}}\)\(= 2 \times 10^{-6} C\) = 2 μC

Therefore, the identical charges are 2μC.

Mistake Points

• One kilogram weight (1 kg wt) is the gravitational force acting on a body of mass 1 kg.
• 1 kg wt is converted to newton as follows: 1 kg \(\times\) g = 1 kg \(\times\) 9.81 m/s² = 9.81 N
  • ​Therefore, 1 g wt = 10^-3 kg \(\times\) 9.81 m/s²   (Assuming g = 9.81 m/s²)

As we know, q = ne,

Where q is the total charge, n is the no of electrons and e is the charge on one electron.

q = 1 mC = 10^-3 C 

e = 1.6 × 10^-19

∴ 10^-3 = n × 1.6 × 10^-19

\( \Rightarrow n = \frac{{{{10}^{ - 3}}}}{{1.6 \times {{10}^{ - 19}}}} = 6.25 \times {10^{15}}\)

CONCEPT:

• Coulomb’s law: When two charged particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.

Force (F) ∝ q1 × q2

\(F \propto \frac{1}{{{r^2}}}\)

\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)

Where K is a constant = 9 × 109 Nm2/C2

CALCULATION:

Here r = d, q₁ = q₂ = e = charge on an electron

As per coulomb's law, force is inversely proportional to the square of the distance.

\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)

\(F = K\frac{{{e} \times {e}}}{{{d^2}}} = Ke^2/d^2\)

So option 3 is correct.

CONCEPT:

• Coulomb law: If two stationary and point charges Q₁ and Q₂ are kept at a distance r, then it is found that force of attraction or repulsion between them is mathematical, Force between point charges:

\(\vec F = k\frac{{{Q_1}{Q_2}}}{{{r^2}}}\)

Where, F = force between them, k = proportionality constant = 9 × 10⁹ Nm²/C², r = distance between charges.

• Opposite charges attract each other
• Same charges repel each other
• When charges are brought in contact, they share charges in an equal ratio.

EXPLANATION:

Let one object receive q and another object receive (Q – q) of charge.

The force between the objects is \(F\; = \;\frac{{q\left( {Q - q} \right)}}{{4\pi {\epsilon_O}{r^2}}}\)

Where r is the distance between the charges,

Now for F to be maximum

\(\frac{{dF}}{{dq}}\; = \;0 \Rightarrow \frac{1}{{4\pi {\epsilon_O}{r^2}}} \times \frac{d}{{dq}}\left( {Qq - {q^2}} \right)\; = \;0\;\)

This means \(q\; = \;\frac{Q}{2}\)

Thus for maximum force charge must be divided equally onto two objects.

So option 1 is correct.