UHS Punjab MDCAT Physics Electrostatics — Set 2

UHS Punjab MDCAT Physics Electrostatics — Set 2

1. Question 1

   Q1. The electric field due to a point charge of 4 μC at a distance of 0.2 m is

   • A) 9 × 10^5 N/C
   • B) 18 × 10^5 N/C
   • C) 36 × 10^5 N/C
   • D) 72 × 10^5 N/C

   Answer: 9 × 10^5 N/C

   Explanation: E = k * q / r^2 = 9 × 10^5 N/C. Wrong options result from miscalculating k or q.

2. Question 2

   Q2. A capacitor has a capacitance of 5 μF when the dielectric is air. If the dielectric is replaced with a material of dielectric constant 4, its capacitance becomes

   • A) 10 μF
   • B) 20 μF
   • C) 5 μF
   • D) 2.5 μF

   Answer: 20 μF

   Explanation: C = ε_r * C0 = 4 * 5 = 20 μF. Wrong options misinterpret dielectric constant effect.

3. Question 3

   Q3. Two parallel plates, each of area 0.02 m^2, are 0.005 m apart. What is the capacitance if the dielectric is air?

   • A) 35.4 pF
   • B) 17.7 pF
   • C) 8.85 pF
   • D) 4.42 pF

   Answer: 35.4 pF

   Explanation: C = ε0 * A / d = 35.4 pF. Wrong options result from incorrect unit conversions or formula application.

4. Question 4

   Q4. A charge of 0.5 C is moved from a point of potential 10 V to another point of potential 20 V. The work done is

   • A) 5 J
   • B) 10 J
   • C) 15 J
   • D) 20 J

   Answer: 5 J

   Explanation: W = q * ΔV = 0.5 * (20 - 10) = 5 J. Wrong options miscalculate ΔV or q.

5. Question 5

   Q5. The potential at a point due to a point charge of 2 μC placed 0.3 m away is

   • A) 6 × 10^4 V
   • B) 3 × 10^4 V
   • C) 12 × 10^4 V
   • D) 9 × 10^4 V

   Answer: 6 × 10^4 V

   Explanation: V = k * q / r = 6 × 10^4 V. Wrong options miscalculate k or r.

6. Question 6

   Q6. The electric flux through a spherical surface of radius 0.1 m enclosing a charge of 1 μC is

   • A) 1.13 × 10^5 Nm^2/C
   • B) 2.26 × 10^5 Nm^2/C
   • C) 0.56 × 10^5 Nm^2/C
   • D) 4.52 × 10^5 Nm^2/C

   Answer: 1.13 × 10^5 Nm^2/C

   Explanation: Φ = q / ε0 = 1.13 × 10^5 Nm^2/C. Wrong options result from incorrect ε0 value.

7. Question 7

   Q7. A parallel plate capacitor is charged to 100 V and then isolated. If the plate separation is doubled, the potential difference becomes

   • A) 50 V
   • B) 100 V
   • C) 200 V
   • D) 400 V

   Answer: 200 V

   Explanation: V = Qd / ε0A, so doubling d doubles V. Wrong options misunderstand capacitor isolation.

8. Question 8

   Q8. A charge of 8 μC is placed at the center of a cube. The total electric flux through the cube is

   • A) 9.04 × 10^5 Nm^2/C
   • B) 4.52 × 10^5 Nm^2/C
   • C) 1.81 × 10^6 Nm^2/C
   • D) 3.62 × 10^6 Nm^2/C

   Answer: 9.04 × 10^5 Nm^2/C

   Explanation: Φ = q / ε0 = 9.04 × 10^5 Nm^2/C. Wrong options misapply Gauss's Law.

9. Question 9

   Q9. The energy stored in a capacitor of capacitance 10 μF charged to 50 V is

   • A) 12.5 mJ
   • B) 25 mJ
   • C) 50 mJ
   • D) 100 mJ

   Answer: 12.5 mJ

   Explanation: E = 0.5 * C * V^2 = 12.5 mJ. Wrong options miscalculate C or V^2.

10. Question 10

    Q10. When a dielectric slab is inserted between the plates of a capacitor, the energy stored

    • A) Increases
    • B) Decreases
    • C) Remains the same
    • D) Becomes zero

    Answer: Increases

    Explanation: E = 0.5 * C * V^2, and C increases with dielectric, so E increases if V is constant. Wrong options misinterpret effect of dielectric.

11. Question 11

    Q11. The electric potential at a distance of 0.2 m from a point charge is 900 V. The charge is

    • A) 20 nC
    • B) 40 nC
    • C) 10 nC
    • D) 80 nC

    Answer: 20 nC

    Explanation: q = 4πε0 * r * V = 20 nC. Wrong options miscalculate ε0 or r.

12. Question 12

    Q12. A capacitor is charged to a potential difference of 200 V. If the charge stored is 0.02 C, the capacitance is

    • A) 100 μF
    • B) 50 μF
    • C) 200 μF
    • D) 400 μF

    Answer: 100 μF

    Explanation: C = Q / V = 0.02 / 200 = 100 μF. Wrong options miscalculate Q or V.

13. Question 13

    Q13. The force between two point charges is 0.036 N when they are 0.2 m apart. If the distance is increased to 0.6 m, the force becomes

    • A) 0.004 N
    • B) 0.012 N
    • C) 0.036 N
    • D) 0.108 N

    Answer: 0.004 N

    Explanation: F ∝ 1/r^2, so F at 0.6 m is (0.2/0.6)^2 times F at 0.2 m = 0.004 N. Wrong options misapply inverse square law.

14. Question 14

    Q14. The electric field between two parallel plates 0.01 m apart is 1000 N/C. The potential difference between the plates is

    • A) 10 V
    • B) 5 V
    • C) 20 V
    • D) 15 V

    Answer: 10 V

    Explanation: V = E * d = 1000 * 0.01 = 10 V. Wrong options miscalculate E or d.

15. Question 15

    Q15. A spherical conductor of radius 0.1 m is charged to a potential of 1000 V. The charge on the conductor is

    • A) 11.1 nC
    • B) 22.2 nC
    • C) 5.55 nC
    • D) 44.4 nC

    Answer: 11.1 nC

    Explanation: q = 4πε0 * r * V = 11.1 nC. Wrong options miscalculate ε0 or r.

16. Question 16

    Q16. A capacitor of capacitance 2 μF is connected in series with another of 3 μF. The equivalent capacitance is

    • A) 1.2 μF
    • B) 1.5 μF
    • C) 0.6 μF
    • D) 5 μF

    Answer: 1.2 μF

    Explanation: 1/C_eq = 1/C1 + 1/C2 gives C_eq = 1.2 μF. Wrong options misapply series capacitance formula.

17. Question 17

    Q17. The capacitance of a parallel plate capacitor is 5 pF. If the area of the plates is doubled, the new capacitance is

    • A) 10 pF
    • B) 20 pF
    • C) 5 pF
    • D) 2.5 pF

    Answer: 10 pF

    Explanation: C ∝ A, so doubling A doubles C to 10 pF. Wrong options misunderstand effect of area on capacitance.

18. Question 18

    Q18. A charge of 2 μC experiences a force of 0.04 N in an electric field. The electric field strength is

    • A) 20000 N/C
    • B) 10000 N/C
    • C) 40000 N/C
    • D) 80000 N/C

    Answer: 20000 N/C

    Explanation: E = F / q = 0.04 / (2 × 10^-6) = 20000 N/C. Wrong options miscalculate F or q.

19. Question 19

    Q19. The potential difference between two points 0.1 m apart in a uniform electric field of 500 N/C is

    • A) 50 V
    • B) 25 V
    • C) 100 V
    • D) 200 V

    Answer: 50 V

    Explanation: ΔV = E * d = 500 * 0.1 = 50 V. Wrong options miscalculate E or d.

20. Question 20

    Q20. Two charges, +4 μC and -6 μC, are separated by 3 meters. What is the electric force between them? (k = 9×10⁹ N·m²/C²)

    • A) 36×10⁻³ N (attraction)
    • B) 24×10⁻³ N (repulsion)
    • C) 12×10⁻³ N (attraction)
    • D) 6×10⁻³ N (repulsion)

    Answer: 36×10⁻³ N (attraction)

    Explanation: Coulomb's Law F=k|q1q2|/r². Option C fails because it uses k=12×10⁹ instead of 9×10⁹.