A Level Physics 18.4 Electric Field of a Point Charge MCQs – 30 Practice Questions with Answers (Cambridge 9702)

This page covers Section 18.4 of the Cambridge International AS & A Level Physics (9702) syllabus, an A Level (A2) topic on the electric field produced by a point charge. Work through 30 original practice MCQs, including field calculations and combining fields from two charges, and check the worked answer after each question.

Formula recap

Field of a point charge: E = Q / (4πε0 r2), where r is measured from the charge. Force on a charge q placed in the field: F = qE. In all calculations on this page, 1 / (4πε0) = 9.0 × 109 m/F. See the full A Level Physics Formula Sheet for the complete Topic 18 equations.

Field of a Point Charge

Q1. Which expression gives the magnitude of the electric field strength at a distance r from a point charge Q?

  • (A) Q / (4πε0 r)
  • (B) Q / (4πε0 r2)
  • (C) Q r2 / (4πε0)
  • (D) 4πε0 Q / r2
Show Answer

Answer: B – The field of a point charge follows an inverse-square law: E = Q / (4πε0 r2). The distance must be squared, and ε0 belongs in the denominator.

Q2. The electric field around an isolated positive point charge points

  • (A) radially outwards from the charge
  • (B) radially inwards towards the charge
  • (C) in circles around the charge
  • (D) along parallel straight lines
Show Answer

Answer: A – Field direction is defined as the direction of the force on a positive test charge. A positive test charge is repelled by a positive point charge, so the field points radially outwards in all directions.

Q3. The electric field around an isolated negative point charge points

  • (A) radially outwards from the charge
  • (B) in circles around the charge
  • (C) radially inwards towards the charge
  • (D) in no direction, because the field is zero
Show Answer

Answer: C – A positive test charge is attracted towards a negative point charge, so the field lines point radially inwards, towards the charge.

Q4. Electric field strength is a vector quantity. When two point charges are present, the total field at a point is found by

  • (A) adding the magnitudes of the two fields
  • (B) subtracting the smaller field magnitude from the larger
  • (C) multiplying the two field magnitudes
  • (D) adding the two fields as vectors, taking their directions into account
Show Answer

Answer: D – Because E is a vector, the fields from separate charges must be combined by vector addition. Magnitudes can only be added or subtracted directly when the fields are parallel or antiparallel.

Q5. A test charge q is placed at a point where the field strength due to a point charge is E. The test charge is replaced by a charge of 2q at the same point. The field strength at that point (due to the source charge) is now

  • (A) unchanged, still E
  • (B) 2E
  • (C) E / 2
  • (D) 4E
Show Answer

Answer: A – E = Q / (4πε0 r2) depends only on the source charge and the distance. Doubling the test charge doubles the force on it, so F / q, and hence E, is unchanged.

Q6. The unit of electric field strength is the

  • (A) newton
  • (B) coulomb per newton
  • (C) newton per coulomb
  • (D) joule per coulomb
Show Answer

Answer: C – E is defined as force per unit charge, E = F / q, so its unit is N C−1. The joule per coulomb is the volt, the unit of potential.

Q7. The electric field of a point charge is similar to the gravitational field of a point mass because both

  • (A) can be repulsive
  • (B) decrease in magnitude with the square of the distance
  • (C) act on uncharged objects
  • (D) are independent of distance
Show Answer

Answer: B – Both fields follow inverse-square laws: E ∝ 1 / r2 and g ∝ 1 / r2. The key difference is that the gravitational field is always attractive, while the electric field direction depends on the sign of the source charge and the force can be attractive or repulsive.

Q8. The electric field strength E at a point is defined as

  • (A) the work done per unit charge in moving a charge to the point
  • (B) the charge stored per unit area at the point
  • (C) the potential energy per unit charge at the point
  • (D) the force acting per unit positive charge placed at the point
Show Answer

Answer: D – E = F / q for a positive test charge q placed at the point. Work done per unit charge describes electric potential, not field strength.

Q9. A charge of −2q is placed at a point where the electric field strength is E, directed to the right. The force on this charge is

  • (A) 2qE, directed to the right
  • (B) zero
  • (C) 2qE, directed to the left
  • (D) qE, directed to the left
Show Answer

Answer: C – F = qE gives a magnitude of 2qE. Because the charge is negative, the force acts opposite to the field direction, so it is directed to the left.

Q10. As the distance r from a point charge increases, the magnitude of the electric field strength

  • (A) decreases as 1 / r2
  • (B) decreases as 1 / r
  • (C) stays constant
  • (D) increases as r2
Show Answer

Answer: A – E = Q / (4πε0 r2), so doubling the distance reduces the field to one quarter. This is why field lines spread out and become less crowded as r increases.

Point-Charge Field Calculations

Q11. Calculate the electric field strength at a distance of 0.30 m from a point charge of +1.0 μC.

  • (A) 1.0 × 104 N/C
  • (B) 1.0 × 105 N/C
  • (C) 3.0 × 105 N/C
  • (D) 9.0 × 105 N/C
Show Answer

Answer: B – E = 9.0 × 109 × 1.0 × 10−6 / (0.30)2 = 9.0 × 103 / 0.090 = 1.0 × 105 N/C.

Q12. Calculate the electric field strength at a distance of 0.20 m from a point charge of +4.0 μC.

  • (A) 9.0 × 104 N/C
  • (B) 1.8 × 105 N/C
  • (C) 9.0 × 105 N/C
  • (D) 3.6 × 106 N/C
Show Answer

Answer: C – E = 9.0 × 109 × 4.0 × 10−6 / (0.20)2 = 3.6 × 104 / 0.040 = 9.0 × 105 N/C.

Q13. At a distance r from a point charge, the field strength is E. What is the field strength at a distance of 2r from the charge?

  • (A) 4E
  • (B) 2E
  • (C) E / 2
  • (D) E / 4
Show Answer

Answer: D – E ∝ 1 / r2, so doubling the distance gives E / 22 = E / 4.

Q14. At a distance r from a point charge, the field strength is E. What is the field strength at a distance of r / 3 from the charge?

  • (A) 9E
  • (B) 3E
  • (C) E / 3
  • (D) E / 9
Show Answer

Answer: A – Reducing the distance to one third multiplies the field by 32 = 9, because E ∝ 1 / r2.

Q15. The electric field strength at a distance of 0.60 m from a point charge is 2.5 × 104 N/C. What is the magnitude of the charge?

  • (A) 0.10 μC
  • (B) 1.0 μC
  • (C) 10 μC
  • (D) 0.50 μC
Show Answer

Answer: B – Rearranging E = Q / (4πε0 r2): Q = E r2 / (9.0 × 109) = 2.5 × 104 × 0.36 / (9.0 × 109) = 9.0 × 103 / 9.0 × 109 = 1.0 × 10−6 C = 1.0 μC.

Q16. A point charge of +1.0 μC produces a field strength of 9.0 × 103 N/C at a point P. How far is P from the charge?

  • (A) 0.10 m
  • (B) 0.30 m
  • (C) 1.0 m
  • (D) 3.0 m
Show Answer

Answer: C – r = √(9.0 × 109 × 1.0 × 10−6 / 9.0 × 103) = √(9.0 × 103 / 9.0 × 103) = √1.0 = 1.0 m.

Q17. A test charge of +3.0 nC is placed 0.15 m from a point charge of +2.0 μC. What force acts on the test charge?

  • (A) 2.4 × 10−3 N
  • (B) 2.4 × 10−4 N
  • (C) 8.0 × 10−3 N
  • (D) 6.0 × 10−4 N
Show Answer

Answer: A – First find the field: E = 9.0 × 109 × 2.0 × 10−6 / (0.15)2 = 1.8 × 104 / 0.0225 = 8.0 × 105 N/C. Then F = qE = 3.0 × 10−9 × 8.0 × 105 = 2.4 × 10−3 N.

Q18. A sphere of radius 0.10 m carries a charge of +5.0 μC, spread over its surface. What is the field strength at the surface of the sphere?

  • (A) 4.5 × 105 N/C
  • (B) 5.0 × 106 N/C
  • (C) 4.5 × 107 N/C
  • (D) 4.5 × 106 N/C
Show Answer

Answer: D – Outside a charged sphere, the field is the same as that of a point charge at the centre, so use r = radius = 0.10 m: E = 9.0 × 109 × 5.0 × 10−6 / (0.10)2 = 4.5 × 104 / 0.010 = 4.5 × 106 N/C.

Q19. What is the field strength at a point 0.50 m from a point charge of −2.0 μC?

  • (A) 7.2 × 104 N/C, directed towards the charge
  • (B) 7.2 × 104 N/C, directed away from the charge
  • (C) 3.6 × 104 N/C, directed towards the charge
  • (D) 1.8 × 105 N/C, directed towards the charge
Show Answer

Answer: A – Magnitude: E = 9.0 × 109 × 2.0 × 10−6 / (0.50)2 = 1.8 × 104 / 0.25 = 7.2 × 104 N/C. The source charge is negative, so the field points towards it.

Q20. A point charge produces a field E at a distance r. The charge is doubled and the distance is also doubled. The new field strength is

  • (A) E / 4
  • (B) E / 2
  • (C) E
  • (D) 2E
Show Answer

Answer: B – E ∝ Q / r2, so the change is a factor of 2 / 22 = 1 / 2. The new field is E / 2.


Combining Fields from Two Charges

Q21. Two point charges, +4.0 μC and +1.0 μC, are placed 0.90 m apart. At what distance from the +4.0 μC charge, along the line joining them, is the resultant field strength zero?

  • (A) 0.30 m
  • (B) 0.45 m
  • (C) 0.60 m
  • (D) 0.90 m
Show Answer

Answer: C – At the null point the two field magnitudes are equal: 4.0 / x2 = 1.0 / (0.90 − x)2. Taking square roots: 2 / x = 1 / (0.90 − x), so 2(0.90 − x) = x, giving x = 0.60 m. The null point lies closer to the smaller charge, as expected.

Q22. A charge of +2.0 μC and a charge of −2.0 μC are separated by 0.40 m. What is the resultant field strength at the midpoint between them?

  • (A) zero
  • (B) 9.0 × 105 N/C, directed towards the negative charge
  • (C) 4.5 × 105 N/C, directed towards the negative charge
  • (D) 9.0 × 105 N/C, directed towards the positive charge
Show Answer

Answer: B – The midpoint is 0.20 m from each charge, so each field has magnitude 9.0 × 109 × 2.0 × 10−6 / 0.040 = 4.5 × 105 N/C. At the midpoint both fields point the same way (away from the positive charge and towards the negative charge), so they add: 2 × 4.5 × 105 = 9.0 × 105 N/C, towards the negative charge.

Q23. A charge of +2.0 μC and a second charge of +2.0 μC are separated by 0.40 m. What is the resultant field strength at the midpoint between them?

  • (A) zero
  • (B) 4.5 × 105 N/C
  • (C) 9.0 × 105 N/C
  • (D) 1.8 × 106 N/C
Show Answer

Answer: A – At the midpoint the two fields have equal magnitudes (4.5 × 105 N/C) but opposite directions, because each points away from its own positive charge. They cancel, so the resultant field is zero.

Q24. A charge of +9.0 μC is at the origin and a charge of +4.0 μC is 0.50 m away from it on the x-axis. What is the resultant field strength at a point on the x-axis 1.0 m from the +9.0 μC charge and 0.50 m beyond the +4.0 μC charge?

  • (A) 6.3 × 104 N/C
  • (B) 1.4 × 105 N/C
  • (C) 3.6 × 105 N/C
  • (D) 2.25 × 105 N/C
Show Answer

Answer: D – Field from the +9.0 μC charge: 9.0 × 109 × 9.0 × 10−6 / 1.02 = 8.1 × 104 N/C. Field from the +4.0 μC charge: 9.0 × 109 × 4.0 × 10−6 / 0.25 = 1.44 × 105 N/C. The point is beyond both charges, so both fields point in the same direction (away from both charges) and add: 0.81 × 105 + 1.44 × 105 = 2.25 × 105 N/C.

Q25. Point P is 0.30 m to the right of a charge of +0.60 μC, and 0.30 m directly above a charge of +0.80 μC. What is the magnitude of the resultant field strength at P?

  • (A) 2.0 × 104 N/C
  • (B) 1.4 × 105 N/C
  • (C) 2.4 × 105 N/C
  • (D) 1.0 × 105 N/C
Show Answer

Answer: D – Field from the +0.60 μC charge at P: 9.0 × 109 × 0.60 × 10−6 / 0.090 = 6.0 × 104 N/C, pointing to the right. Field from the +0.80 μC charge: 9.0 × 109 × 0.80 × 10−6 / 0.090 = 8.0 × 104 N/C, pointing upwards. These are perpendicular, so by Pythagoras the resultant is √((6.0 × 104)2 + (8.0 × 104)2) = √(1.0 × 1010) = 1.0 × 105 N/C.

Q26. Two point charges, +Q and −4Q, are held a fixed distance apart. The point where the resultant field strength is zero is located

  • (A) between the charges, closer to the +Q charge
  • (B) between the charges, closer to the −4Q charge
  • (C) beyond the −4Q charge, on the side away from the +Q charge
  • (D) beyond the +Q charge, on the side away from the −4Q charge
Show Answer

Answer: D – Between the charges the two fields point in the same direction, so they cannot cancel there. Outside the pair, the fields oppose each other, and they can only balance where the weaker but nearer charge (+Q, the smaller magnitude) is closer to the point: beyond the +Q charge.

Q27. Charges of +2.0 μC and +8.0 μC are separated by 0.60 m. What is the resultant field strength at the midpoint between them?

  • (A) 6.0 × 105 N/C, directed towards the +2.0 μC charge
  • (B) 6.0 × 105 N/C, directed towards the +8.0 μC charge
  • (C) 1.0 × 106 N/C, directed towards the +2.0 μC charge
  • (D) zero
Show Answer

Answer: A – At the midpoint (0.30 m from each): field from +2.0 μC is 9.0 × 109 × 2.0 × 10−6 / 0.090 = 2.0 × 105 N/C pointing towards the +8.0 μC charge; field from +8.0 μC is 9.0 × 109 × 8.0 × 10−6 / 0.090 = 8.0 × 105 N/C pointing towards the +2.0 μC charge. They oppose, so the resultant is 8.0 × 105 − 2.0 × 105 = 6.0 × 105 N/C, directed towards the +2.0 μC charge.

Q28. A charge of +6.0 μC and a charge of +2.0 μC are 0.40 m apart. Point P lies on the line joining them, 0.20 m beyond the +2.0 μC charge (so P is 0.60 m from the +6.0 μC charge). What is the resultant field strength at P?

  • (A) 1.5 × 105 N/C
  • (B) 3.0 × 105 N/C
  • (C) 6.0 × 105 N/C
  • (D) 4.5 × 105 N/C
Show Answer

Answer: C – Field from +6.0 μC: 9.0 × 109 × 6.0 × 10−6 / 0.36 = 1.5 × 105 N/C. Field from +2.0 μC: 9.0 × 109 × 2.0 × 10−6 / 0.040 = 4.5 × 105 N/C. P is beyond both charges, so both fields point away in the same direction and add: 1.5 × 105 + 4.5 × 105 = 6.0 × 105 N/C.

Q29. Charges of +9.0 μC and −4.0 μC are held 0.50 m apart. Where is the point at which the resultant field strength is zero?

  • (A) At the midpoint between the charges
  • (B) Between the charges, 0.20 m from the −4.0 μC charge
  • (C) 1.0 m beyond the −4.0 μC charge
  • (D) 1.0 m beyond the +9.0 μC charge
Show Answer

Answer: C – The null point must be outside the pair, on the side of the smaller-magnitude charge. Let it be x beyond the −4.0 μC charge: 9.0 / (0.50 + x)2 = 4.0 / x2. Taking square roots: 3 / (0.50 + x) = 2 / x, so 3x = 1.0 + 2x and x = 1.0 m beyond the −4.0 μC charge.

Q30. At a point in space, the resultant field from two point charges is 3.0 × 105 N/C, directed to the right. A small charge of −2.0 nC is placed at this point. What force acts on it?

  • (A) 6.0 × 10−4 N, directed to the right
  • (B) 6.0 × 10−4 N, directed to the left
  • (C) 1.5 × 10−4 N, directed to the left
  • (D) 6.0 × 10−3 N, directed to the left
Show Answer

Answer: B – F = qE = 2.0 × 10−9 × 3.0 × 105 = 6.0 × 10−4 N. The charge is negative, so the force is opposite to the field direction: to the left.

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