Definition
Electric field
Force per unit positive test charge at a point: E = F/q. SI unit: N/C or V/m. Field due to point charge: E = (1/4πε₀) q/r² along the radial direction.
-- NCERT Class 12 Physics, Ch. 1, p. 15Electric Field Point Charge
8 MCQs1 revision card9-step worked example
Source: NCERT ElectrostaticsOfficial key: NTA-verifiedLast updated: 21 Sep 2026
Electric Field Point Charge, explained for NEET
The trap: treating the field as a property of the test charge. Students double the test charge and double E, or point the force on an electron along E. NCERT Class 12 Physics Part I, Chapter 1 (Section 1.7, page 15) says the field due to Q is independent of the test charge q: F is proportional to q, so F/q does not depend on q.
The field. A point charge Q in vacuum produces an electric field at every point around it (NCERT Chapter 1, Eq. 1.6, page 14):
E = (1/4πε₀) Q/r², directed along r̂, the unit vector from the charge to the point.
A charge q placed where the field is E feels F = qE (Eq. 1.8, page 14). This equation defines the SI unit of electric field as N/C; NCERT notes that an alternate unit, V/m, comes in the next chapter. Because Q itself would be pushed by q, the test charge is made negligibly small: E is the limit of F/q as q → 0 (Eq. 1.9, page 15). Equivalently, E at a point is the force a unit positive charge would experience there.
Direction and symmetry (page 15). For a positive source charge the field points radially outwards; for a negative source charge it points radially inwards. The magnitude depends only on r, so it is the same at every point of a sphere centred on the charge.
Bridge to NEET. Typical items give Q and r and ask for E, give E and q and ask for F, or give E and r and ask for Q. Distractors come from using r instead of r², dropping 1/4πε₀, inverting F/q, or scaling E as 1/r when r changes.
Watch out: F = qE carries the sign of q. On a negative charge the force is opposite to E; on a positive charge it is along E. The field itself does not change when you swap the test charge.
Can you answer these Electric Field Point Charge MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
From the relation F = qE, the SI unit of electric field is:
Show answer and why every option is right or wrong
Answer: B. E = F/q, so its unit is newton per coulomb, N/C (NCERT Class 12 Physics Part I, Chapter 1, Eq. 1.8, page 14).
Why A is wrong: A is wrong because it multiplies force by charge; F = qE gives E = F/q, a ratio, not a product.
Why C is wrong: C is wrong because it inverts the ratio (q/F instead of F/q); C/N is the unit of 1/E.
Why D is wrong: D is wrong because N·m²/C² is the unit of the constant 1/4πε₀, not of the field.
MCQ 2Easy RecallPractice
The electric field produced by an isolated negative point charge, at points around it, is directed:
Show answer and why every option is right or wrong
Answer: A. NCERT Class 12 Physics Part I, Chapter 1, Section 1.7 (page 15): if the source charge is negative, the field vector at each point points radially inwards; for a positive source charge it points radially outwards.
Why B is wrong: B is wrong because the field of a point charge is along r̂, the line from the charge to the point, not perpendicular to it.
Why C is wrong: C is wrong because radially outward is the direction for a positive source charge; a negative source charge gives an inward field.
Why D is wrong: D is wrong because the field direction is set by the sign of the source charge; the test charge only changes the direction of the force F = qE, not of E.
MCQ 3Easy RecallPractice
A small test charge placed at point P, near a fixed point charge Q, experiences a force. The test charge is replaced by one of twice the charge. The electric field due to Q at P:
Show answer and why every option is right or wrong
Answer: C. NCERT Class 12 Physics Part I, Chapter 1, Section 1.7 (page 15): E due to Q is independent of q, because F is proportional to q and the ratio F/q does not depend on q. The force doubles; the field does not change.
Why A is wrong: A is wrong because it is the force F = qE that doubles; the field E = F/q stays the same since F and q both double.
Why B is wrong: B is wrong because it divides the unchanged force by the doubled charge; the force also doubles, so F/q is unchanged.
Why D is wrong: D is wrong because it treats the field as depending on the square of the test charge; the field due to Q does not depend on the test charge at all.
MCQ 4Direct ApplicationPractice
What is the magnitude of the electric field at a distance of 0.30 m from a point charge of +2.0 × 10⁻⁹ C in vacuum? (Take 1/4πε₀ = 9.0 × 10⁹ N m² C⁻².)
Show answer and why every option is right or wrong
Answer: A. E = (1/4πε₀) Q/r² = (9.0 × 10⁹)(2.0 × 10⁻⁹)/(0.30)² = 18/0.090 = 2.0 × 10² N/C (NCERT Class 12 Physics Part I, Chapter 1, Eq. 1.6, page 14).
Why B is wrong: B is wrong because 6.0 × 10¹ N/C = 18/0.30 divides by r instead of r².
Why C is wrong: C is wrong because 3.0 × 10¹ N/C = 18/0.60 doubles r (2 × 0.30) instead of squaring it.
Why D is wrong: D is wrong because 2.2 × 10⁻⁸ N/C = (2.0 × 10⁻⁹)/(0.090) drops the constant 1/4πε₀.
MCQ 5Direct ApplicationPractice
A uniform electric field of magnitude 2.0 × 10⁴ N/C points along the +x direction. What force acts on an electron (charge −1.6 × 10⁻¹⁹ C) placed in it?
Show answer and why every option is right or wrong
Answer: D. F = qE: magnitude (1.6 × 10⁻¹⁹)(2.0 × 10⁴) = 3.2 × 10⁻¹⁵ N. Since q is negative, F is opposite to E, i.e. along −x (NCERT Class 12 Physics Part I, Chapter 1, Eq. 1.8, page 14).
Why A is wrong: A is wrong because 1.3 × 10²³ = (2.0 × 10⁴)/(1.6 × 10⁻¹⁹) = 1.25 × 10²³ divides the field by the charge instead of multiplying.
Why B is wrong: B is wrong because the magnitude is right but it ignores the negative sign of the electron's charge; for negative q the force is opposite to E.
Why C is wrong: C is wrong because 8.0 × 10⁻²⁴ = (1.6 × 10⁻¹⁹)/(2.0 × 10⁴) divides the charge by the field instead of multiplying.
MCQ 6Direct ApplicationPractice
A test charge of +2.0 × 10⁻⁶ C placed at point P experiences an electric force of 8.0 × 10⁻³ N. The magnitude of the electric field at P is:
Show answer and why every option is right or wrong
Answer: C. E = F/q = (8.0 × 10⁻³ N)/(2.0 × 10⁻⁶ C) = 4.0 × 10³ N/C (NCERT Class 12 Physics Part I, Chapter 1, Eqs. 1.8 and 1.9, pages 14–15).
Why A is wrong: A is wrong because 1.6 × 10⁻⁸ = (8.0 × 10⁻³)(2.0 × 10⁻⁶) multiplies force and charge instead of dividing.
Why B is wrong: B is wrong because 2.5 × 10⁻⁴ = (2.0 × 10⁻⁶)/(8.0 × 10⁻³) inverts the ratio, giving q/F instead of F/q.
Why D is wrong: D is wrong because NCERT states the field is produced by the source charge everywhere around it and is independent of the test charge; the test charge only detects it.
MCQ 7CalculationPractice
At a distance of 0.20 m from an isolated point charge, the electric field has magnitude 9.0 × 10³ N/C. At a distance of 0.60 m from the same charge, what are the field magnitude and the magnitude of the force on a +1.0 × 10⁻⁶ C charge placed there?
Show answer and why every option is right or wrong
Answer: B. E ∝ 1/r². Tripling r (0.20 m → 0.60 m) divides E by 3² = 9: 9.0 × 10³/9 = 1.0 × 10³ N/C. Then F = qE = (1.0 × 10⁻⁶)(1.0 × 10³) = 1.0 × 10⁻³ N (NCERT Class 12 Physics Part I, Chapter 1, Eqs. 1.6 and 1.8, page 14).
Why A is wrong: A is wrong because 3.0 × 10³ N/C = 9.0 × 10³/3 scales the field as 1/r instead of 1/r²; the force 3.0 × 10⁻³ N then carries the same error.
Why C is wrong: C is wrong because it keeps the field unchanged at 9.0 × 10³ N/C; NCERT states the magnitude depends on r, so moving farther away changes it.
Why D is wrong: D is wrong because 8.1 × 10⁴ N/C = 9.0 × 10³ × 9 multiplies by 9, as if the field grew with distance; it falls as 1/r².
MCQ 8CalculationPractice
At a point 0.20 m from an isolated point charge Q, the electric field has magnitude 4.5 × 10⁵ N/C and points towards Q. (Take 1/4πε₀ = 9.0 × 10⁹ N m² C⁻².) The charge Q is:
Show answer and why every option is right or wrong
Answer: D. |Q| = E r² × 4πε₀ = (4.5 × 10⁵)(0.20)²/(9.0 × 10⁹) = (4.5 × 10⁵)(0.040)/(9.0 × 10⁹) = 2.0 × 10⁻⁶ C. A field pointing radially inwards means the source charge is negative (NCERT Class 12 Physics Part I, Chapter 1, Eq. 1.6, page 14; Section 1.7, page 15). So Q = −2.0 × 10⁻⁶ C.
Why A is wrong: A is wrong because 2.0 × 10⁻⁵ = (4.5 × 10⁵)(0.40)/(9.0 × 10⁹) takes (0.20)² as 0.40, doubling r instead of squaring it.
Why B is wrong: B is wrong because the magnitude is right but the sign is not; a field pointing towards the charge (radially inwards) comes from a negative source charge.
Why C is wrong: C is wrong because 1.0 × 10⁻⁵ = (4.5 × 10⁵)(0.20)/(9.0 × 10⁹) uses r instead of r².
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Electric Field Point Charge: quick recall before you leave
How do you solve a Electric Field Point Charge question? A worked example
- 1
Given
NCERT Class 12 Physics Part I, Chapter 1, Example 1.8 (pages 18–19). Two point charges q₁ = +10⁻⁸ C and q₂ = −10⁻⁸ C are placed 0.1 m apart. Reading the distances used in NCERT's solution (Fig. 1.11): point A is the midpoint, 0.05 m from each charge; point B lies on the line of the charges, 0.05 m from q₁ on the side away from q₂ (so 0.15 m from q₂); point C is 0.10 m from each charge. Take 1/4πε₀ = 9 × 10⁹ N m² C⁻².
- 2
Required
The electric field (magnitude and direction) at A, B and C.
- 3
Concept
Find each charge's field with E = (1/4πε₀)|q|/r². The field of positive q₁ points away from q₁; the field of negative q₂ points towards q₂. Then add the two field vectors at each point. Take "right" as the direction from q₁ towards q₂, as in NCERT's figure.
- 4
Formula
E = (1/4πε₀) |q|/r² (Eq. 1.6); the net field at a point is the vector sum of the two fields.
- 5
Substitution
• At A: E₁A = E₂A = (9 × 10⁹)(10⁻⁸)/(0.05)²• At B: E₁B = (9 × 10⁹)(10⁻⁸)/(0.05)²; E₂B = (9 × 10⁹)(10⁻⁸)/(0.15)²• At C: E₁C = E₂C = (9 × 10⁹)(10⁻⁸)/(0.10)²; each makes an angle π/3 with the line of the charges
- 6
Calculation
• A: E₁A = 90/0.0025 = 3.6 × 10⁴ N/C, pointing right (away from q₁). E₂A has the same magnitude and also points right (towards q₂). E_A = E₁A + E₂A = 7.2 × 10⁴ N/C.• B: E₁B = 3.6 × 10⁴ N/C, pointing left (away from q₁). E₂B = 90/0.0225 = 4 × 10³ N/C, pointing right (towards q₂). E_B = E₁B − E₂B = 3.2 × 10⁴ N/C.• C: E₁C = E₂C = 90/0.01 = 9 × 10³ N/C. The components perpendicular to the line of the charges cancel; the components along it add: E_C = E₁C cos(π/3) + E₂C cos(π/3) = 9 × 10³ × ½ × 2 = 9 × 10³ N/C.
The factor cos(π/3) = ½ and the count of two equal contributions are exact and do not contribute to the significant-figure count; the answers are reported as NCERT prints them. - 7
Final answer
E_A = 7.2 × 10⁴ N/C towards the right; E_B = 3.2 × 10⁴ N/C towards the left; E_C = 9 × 10³ N/C towards the right (NCERT Example 1.8, pages 18–19).
- 8
Common trap
At A, assuming the fields of equal and opposite charges cancel at the midpoint and writing E_A = 0. The field of the positive charge points away from it and the field of the negative charge points towards it, so between the charges both point the same way and add. At B the opposite slip happens: adding the magnitudes (3.6 × 10⁴ + 4 × 10³ = 4.0 × 10⁴ N/C) when the two fields point in opposite directions and must be subtracted.
- 9
Similar NEET-style question
"Point charges of +3.0 × 10⁻⁹ C and −3.0 × 10⁻⁹ C are held 0.60 m apart in vacuum. Find the magnitude and direction of the electric field at the midpoint. (Take 1/4πε₀ = 9.0 × 10⁹ N m² C⁻².)"
Strategy: Each charge is 0.30 m from the midpoint, so each field is (9.0 × 10⁹)(3.0 × 10⁻⁹)/(0.30)² = 27/0.090 = 3.0 × 10² N/C. Both point towards the negative charge, so they add: 6.0 × 10² N/C, directed towards the negative charge. Do not subtract them to get zero.
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What to remember before solving Electric Field Point Charge questions
Which Electric Field Point Charge formulas do you need for NEET?
1 formula — click to collapse
Electric field of point charge
Field magnitude from point charge q at distance r, directed radially.
| Symbol | Quantity | SI Unit |
|---|---|---|
| E | electric field | N/C or V/m |
| q | point charge | C |
| r | distance | m |
Valid when
- Point or spherically symmetric source
- r > source body radius
More in Electrostatics: 3 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
Electric Field Point Charge questions from past NEET papers
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Sources
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