Coulomb's Law & Electric Field — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Coulomb's Law & Electric Field MCQs with step-by-step solutions (21 questions). Part of Electrostatics. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Coulomb's Law & Electric Field · easy · theory
Coulomb's law states that the force between two point charges is:
A. Inversely proportional to the product of the charges
B. Directly proportional to the separation between them
C. Directly proportional to the product of the charges and inversely proportional to the square of their separation ✓ Correct
D. Independent of the medium between them
Solution: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$, an inverse-square law like gravitation but vastly stronger.
Q2 — Coulomb's Law & Electric Field · easy · theory
The electrostatic force between two point charges acts:
A. At $45^\circ$ to the line joining them
B. Along the line joining the two charges ✓ Correct
C. Perpendicular to the line joining them
D. In a direction that depends on the medium
Solution: The Coulomb force is central: it is directed along the line of centres, repulsive for like charges and attractive for unlike ones.
Q3 — Coulomb's Law & Electric Field · medium · theory
The forces that two point charges exert on each other are:
A. Always attractive
B. Equal in magnitude and opposite in direction ✓ Correct
C. Unequal if the charges are unequal
D. Equal in magnitude and in the same direction
Solution: Coulomb forces obey Newton's third law, so the larger charge does not exert the greater force.
Q4 — Coulomb's Law & Electric Field · medium · theory
The permittivity of free space $\varepsilon_0$ has the value:
A. $9 \times 10^9\text{ C}^2\text{N}^{-1}\text{m}^{-2}$
B. $1.6 \times 10^{-19}\text{ C}^2\text{N}^{-1}\text{m}^{-2}$
C. $8.85 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2}$ ✓ Correct
D. $8.85 \times 10^{-12}\text{ N}\text{m}^2\text{C}^{-2}$
Solution: It is related to the Coulomb constant by $\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N}\text{m}^2\text{C}^{-2}$.
Q5 — Coulomb's Law & Electric Field · easy · theory
The SI unit of electric field intensity is:
A. $\text{N}\cdot\text{C}$
B. $\text{N/C}$, equivalently $\text{V/m}$ ✓ Correct
C. $\text{J/C}$
D. $\text{C/N}$
Solution: Electric field is force per unit charge; the alternative form $\text{V/m}$ follows from $E = -\dfrac{dV}{dr}$.
Q6 — Coulomb's Law & Electric Field · medium · theory
Two electric lines of force can never intersect because:
A. Charge would accumulate at the crossing
B. The field would become infinite there
C. The field would then have two directions at the same point ✓ Correct
D. The lines would become closed curves
Solution: The tangent to a line of force gives the field direction, which must be unique at every point.
Q7 — Coulomb's Law & Electric Field · easy · theory
The electric field inside the body of a charged conductor in electrostatic equilibrium is:
A. Directed radially inward
B. Zero ✓ Correct
C. Equal to the surface field
D. Maximum at the centre
Solution: Free charges redistribute themselves on the surface until the interior field vanishes; otherwise they would keep moving.
Q8 — Coulomb's Law & Electric Field · medium · theory
The force between two charges placed in a medium of dielectric constant $K$ compared with that in vacuum is:
A. Unchanged
B. Increased by a factor $K$
C. Reduced by a factor $K^2$
D. Reduced by a factor $K$ ✓ Correct
Solution: The medium partially screens the charges, so $F_{medium} = \dfrac{F_{vacuum}}{K}$.
Q9 — Coulomb's Law & Electric Field · medium · numerical
Two equal point charges of $+10\,\mu\text{C}$ are placed in vacuum $30\text{ cm}$ apart. The force between them is ($\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9$):
A. $1\text{ N}$
B. $10\text{ N}$ ✓ Correct
C. $0.1\text{ N}$
D. $100\text{ N}$
Solution: $F = \dfrac{9 \times 10^9 \times (10^{-5})^2}{(0.3)^2} = \dfrac{9 \times 10^9 \times 10^{-10}}{0.09} = 10\text{ N}$.
Q10 — Coulomb's Law & Electric Field · medium · numerical
Charges of $2\,\mu\text{C}$ and $3\,\mu\text{C}$ are placed $10\text{ cm}$ apart in vacuum. The force between them is:
A. $0.54\text{ N}$
B. $5.4\text{ N}$ ✓ Correct
C. $54\text{ N}$
D. $1.8\text{ N}$
Solution: $F = \dfrac{9 \times 10^9 \times 2 \times 10^{-6} \times 3 \times 10^{-6}}{(0.1)^2} = \dfrac{0.054}{0.01} = 5.4\text{ N}$.
Q11 — Coulomb's Law & Electric Field · easy · numerical
If the distance between two point charges is doubled, the electrostatic force between them becomes:
A. One-half
B. One-fourth ✓ Correct
C. Twice
D. Four times
Solution: $F \propto \dfrac{1}{r^2}$, so doubling the separation reduces the force to a quarter.
Q12 — Coulomb's Law & Electric Field · easy · numerical
If both charges of a pair are doubled while their separation is unchanged, the force between them becomes:
A. Four times ✓ Correct
B. Twice
C. Unchanged
D. Half
Solution: $F \propto q_1q_2$, so doubling each charge multiplies the force by $4$.
Q13 — Coulomb's Law & Electric Field · medium · numerical
The electric field at a distance of $0.3\text{ m}$ from a point charge of $1\,\mu\text{C}$ is:
A. $9 \times 10^3\text{ N/C}$
B. $3 \times 10^4\text{ N/C}$
C. $10^5\text{ N/C}$ ✓ Correct
D. $10^4\text{ N/C}$
Solution: $E = \dfrac{9 \times 10^9 \times 10^{-6}}{(0.3)^2} = \dfrac{9 \times 10^3}{0.09} = 10^5\text{ N/C}$.
Q14 — Coulomb's Law & Electric Field · easy · numerical
A charge of $2\,\mu\text{C}$ placed at a point experiences a force of $0.02\text{ N}$. The electric field there is:
A. $10^5\text{ N/C}$
B. $4 \times 10^{-8}\text{ N/C}$
C. $10^{-8}\text{ N/C}$
D. $10^4\text{ N/C}$ ✓ Correct
Solution: $E = \dfrac{F}{q} = \dfrac{0.02}{2 \times 10^{-6}} = 10^4\text{ N/C}$.
Q15 — Coulomb's Law & Electric Field · hard · numerical
Two identical conducting spheres carry charges $+6\,\mu\text{C}$ and $-2\,\mu\text{C}$ separated by $r$. They are touched together and replaced at the same separation. The ratio of the new force to the original force is:
A. $4 : 3$
B. $1 : 12$
C. $3 : 1$
D. $1 : 3$ ✓ Correct
Solution: On contact the charge equalises at $\dfrac{6 - 2}{2} = +2\,\mu\text{C}$ each. The product of magnitudes changes from $12$ to $4$, so the ratio is $\dfrac{4}{12} = \dfrac{1}{3}$.
Q16 — Coulomb's Law & Electric Field · hard · numerical
Two charges of $+2\,\mu\text{C}$ and $+6\,\mu\text{C}$ repel each other with a force of $12\text{ N}$. If $-4\,\mu\text{C}$ is added to each, the new force is:
A. $4\text{ N}$, repulsive
B. Zero
C. $4\text{ N}$, attractive ✓ Correct
D. $8\text{ N}$, attractive
Solution: The charges become $-2\,\mu\text{C}$ and $+2\,\mu\text{C}$. The product of magnitudes falls from $12$ to $4$, so $F = 4\text{ N}$, now attractive because the signs are opposite.
Q17 — Coulomb's Law & Electric Field · easy · numerical
The force between two charges in vacuum is $F$. When they are placed in a medium of dielectric constant $2$ at the same separation, the force becomes:
A. $4F$
B. $2F$
C. $\dfrac{F}{2}$ ✓ Correct
D. $\dfrac{F}{4}$
Solution: $F_{medium} = \dfrac{F}{K} = \dfrac{F}{2}$.
Q18 — Coulomb's Law & Electric Field · easy · numerical
A charge of $5\,\mu\text{C}$ is placed in a uniform electric field of $2 \times 10^4\text{ N/C}$. The force on it is:
A. $0.1\text{ N}$ ✓ Correct
B. $0.01\text{ N}$
C. $4 \times 10^9\text{ N}$
D. $1\text{ N}$
Solution: $F = qE = 5 \times 10^{-6} \times 2 \times 10^4 = 0.1\text{ N}$.
Q19 — Coulomb's Law & Electric Field · medium · numerical
The force on an electron placed in a uniform electric field of $10^4\text{ N/C}$ is:
A. $1.6 \times 10^{-15}\text{ N}$ ✓ Correct
B. $1.6 \times 10^{-23}\text{ N}$
C. $1.6 \times 10^{-19}\text{ N}$
D. $1.6 \times 10^{-11}\text{ N}$
Solution: $F = eE = 1.6 \times 10^{-19} \times 10^4 = 1.6 \times 10^{-15}\text{ N}$.
Q20 — Coulomb's Law & Electric Field · medium · numerical
Two equal positive charges are separated by a distance $2a$. The electric field at the midpoint of the line joining them is:
A. $\dfrac{2kq}{a^2}$
B. Maximum
C. $\dfrac{kq}{a^2}$
D. Zero ✓ Correct
Solution: The two fields at the midpoint are equal in magnitude but oppositely directed, so they cancel exactly.
Q21 — Coulomb's Law & Electric Field · hard · numerical
Charges of $4\,\mu\text{C}$ and $9\,\mu\text{C}$ are placed $10\text{ cm}$ apart. The distance of the neutral point from the $4\,\mu\text{C}$ charge is:
A. $3\text{ cm}$
B. $5\text{ cm}$
C. $6\text{ cm}$
D. $4\text{ cm}$ ✓ Correct
Solution: Setting the fields equal: $\dfrac{4}{x^2} = \dfrac{9}{(10-x)^2} \Rightarrow \dfrac{2}{x} = \dfrac{3}{10-x} \Rightarrow 20 - 2x = 3x \Rightarrow x = 4\text{ cm}$.