Every physics question lost on this paper, rebuilt in full. Each one carries what was given, what was asked, the concept behind it, the formula, the steps written out, a table justifying the right option and ruling out each of the others, the fastest route through, and an animated figure wherever seeing the thing settles the answer.
12Questions lost
5Attempted, wrong
7Left blank
53Marks at stake
The shape of this paper
Twelve questions from Electric Charges and Fields: five attempted and missed, seven left
blank. On NEET marking that is 48 marks not gained plus 5 in negatives.
The five wrong answers share one cause, and it is worth naming before reading any card:
electric field is a vector, and it was repeatedly treated as a number. Q99 added two
fields that oppose; Q112 reversed a sign; Q122 split a gap in the wrong ratio; Q105 and Q111
both accepted a false reason about direction of charge transfer.
The seven blanks are a different problem. Four of them — Q113, Q115, Q124 and Q130 are
one-line substitutions, and Q118 collapses to a five-second observation.
All twelve come from one chapter. Five were attempted and missed, seven left blank — and every one of the five turns on a direction or a sign rather than on a formula.
Q 99Attempted · wrongSuperposition · field at a mid-point
Two charges +5 μC and +10 μC are placed 20 cm apart. The net electric field at the mid-point between the two charges is
Correct4.5 × 10⁶ N/C directed towards +5 μC
4.5 × 10⁶ N/C directed towards +10 μC
13.5 × 10⁶ N/C directed towards +5 μC
Her answer13.5 × 10⁶ N/C directed towards +10 μC
Given
q₁ = +5 μC, q₂ = +10 μC.
Separation 20 cm, so the mid-point is 0.10 m from each.
Both charges are positive.
Asked
The magnitude and direction of the net field at the mid-point.
Concept to use
Field is a vector, so the two contributions must be combined with their directions. The field of a positive charge points away from it, so at the mid-point the field from the left charge points right and the field from the right charge points left. They oppose, so the magnitudes subtract, and the survivor points away from whichever charge is bigger.
DiagramAnimatedTwo opposing arrows at the mid-point — so subtract, not add.
Formula to useE = kq/r² , k = 9 × 10⁹ ; opposing fields SUBTRACT
Baby steps
From +5 μC: E₁ = 9×10⁹ × 5×10⁻⁶ / (0.1)² = 4.5 × 10⁶ N/C, pointing towards the +10 μC side.
From +10 μC: E₂ = 9×10⁹ × 10×10⁻⁶ / (0.1)² = 9 × 10⁶ N/C, pointing towards the +5 μC side.
They point in opposite directions, so subtract: 9 − 4.5 = 4.5 × 10⁶ N/C.
The larger field wins, so the net points away from +10 μC, that is, towards +5 μC.
Answer
4.5 × 10⁶ N/C directed towards +5 μC
Why this option and not the others
Option
Verdict
Reason
4.5×10⁶ towards +5μC
keep
Magnitudes subtract because the fields oppose, and the net points away from the bigger charge.
4.5×10⁶ towards +10μC
rule out
Right magnitude, wrong direction. The bigger charge pushes the field away from itself, not towards itself.
13.5×10⁶ towards +5μC
rule out
13.5 = 9 + 4.5, which would require the two fields to point the same way. Between two LIKE charges they always oppose.
13.5×10⁶ towards +10μC
rule out
Wrong on both counts — added instead of subtracted, and the direction reversed as well.
Shortcut
Between two like charges the fields always oppose, so subtract; between two unlike charges they reinforce, so add. Decide that from the signs before touching a calculator, and the direction follows for free: away from the bigger charge.
Where it went wrong
13.5 is 9 + 4.5 — the two fields were added rather than subtracted, and the direction was taken towards the larger charge. Both slips come from the same root: treating field as a number rather than a vector. The habit that fixes it is to draw the two arrows at the field point before computing anything. With both charges positive, the arrows visibly point in opposite directions, and adding stops being tempting.
Q 105Attempted · wrongAttraction without both bodies charged
Assertion (A): If there exists a net coulomb attraction between two bodies both of them need not be charged. Reason (R): For net coulomb attraction, two bodies must be oppositely charged.
Both A and R are true and R is correct explanation of A
Both A and R are true and R is not explaining A
CorrectA is true, R is false
Her answerA is false, R is true
Given
Two bodies showing a net Coulomb attraction.
Asked
The truth of each statement.
Concept to use
A charged body attracts a neutral one by induction. The charged body pulls opposite charge to the near face of the neutral body and pushes like charge to the far face; since the near face is closer, its attraction beats the far face's repulsion and the net effect is attraction. So attraction proves nothing about the second body being charged.
DiagramAnimatedA charged body attracting a neutral one, by induction.
Formula to useattraction is possible with: unlike charges OR one charge + one neutral body
Baby steps
Assertion. A charged rod picks up small bits of paper, which are electrically neutral. So attraction can occur with only one body charged. True.
Reason. It claims the bodies must be oppositely charged. The paper example is an immediate counterexample. False.
Note that the reason actually contradicts the assertion rather than supporting it — a strong hint that they cannot both be true.
So A is true and R is false.
Answer
A is true, R is false
Why this option and not the others
Option
Verdict
Reason
Both true, R explains A
rule out
The reason denies what the assertion states, so they cannot both be true.
Both true, R does not explain
rule out
Same objection.
A true, R false
keep
Induction allows a charged body to attract a neutral one, which makes A true and R false.
A false, R true
rule out
The reverse of the truth. Repulsion, not attraction, is the sure test of charge.
Shortcut
Remember the standard line: repulsion is the sure test of electrification, not attraction. Attraction can be produced by induction; repulsion cannot. That single sentence decides every question of this shape.
And a structural clue: when the reason contradicts the assertion, they can never both be true — so options (a) and (b) are dead before you even judge the physics.
Where it went wrong
The two statements were judged the wrong way round. The reason sounds authoritative because opposite charges do attract — but the word that breaks it is must. One counterexample is enough, and the comb-and-paper demonstration is exactly that. Note the pattern with Q111 on this same paper: both are assertion-reason questions where the assertion is true and the reason is a plausible but false generalisation.
Q 106Attempted · wrongPercentage change in Coulomb force
Electrical force between two point charges is 200 N. If we increase 10% charge on one of the charges and decrease 10% charge on the other, then electrical force between them for the same distance becomes
Correct198 N
100 N
Her answer200 N
99 N
Given
Original force F = 200 N.
One charge increased by 10%, the other decreased by 10%.
The separation is unchanged.
Asked
The new force.
Concept to use
Coulomb's law has the two charges multiplied together, so percentage changes multiply too — they do not add and they do not cancel. A rise of 10% is a factor 1.1 and a fall of 10% is a factor 0.9, and 1.1 × 0.9 = 0.99, not 1.
DiagramAnimated+10% and −10% multiply to 0.99, not to 1.
Formula to useF ∝ q₁q₂ → F' / F = 1.1 × 0.9 = 0.99
Baby steps
F = kq₁q₂/r², and r does not change.
New charges: 1.1q₁ and 0.9q₂.
New product: 1.1 × 0.9 × q₁q₂ = 0.99 q₁q₂.
F' = 0.99 × 200 = 198 N.
Answer
198 N
Why this option and not the others
Option
Verdict
Reason
198 N
keep
0.99 × 200. The 1% loss comes from multiplying 1.1 by 0.9.
100 N
rule out
Half the original, which no combination of ±10% can produce.
200 N
rule out
This assumes the +10% and −10% cancel exactly. They do not, because the charges multiply rather than add.
99 N
rule out
The right factor, 0.99, applied to 100 instead of 200.
Shortcut
Learn the general result and this becomes instant: for a product, +x% and −x% together give a loss of x²/100 per cent. Here x = 10, so the loss is 1%, giving 198 N. It is always a loss, never a gain, and never exactly zero.
Where it went wrong
200 N assumes the two 10% changes cancel. They would, if the charges were added — but Coulomb's law multiplies them, and 1.1 × 0.9 = 0.99 rather than 1.00. This is a general arithmetic fact worth carrying well beyond electrostatics: raising one factor and lowering another by the same percentage always leaves the product slightly smaller.
Q 109Left blankSHM of a charge between two fixed charges
Two electrons each are fixed at a distance 2d. A third charge proton placed at the midpoint is displaced slightly by a distance x (x ≪ d) perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency:
(2πε₀md³ / q²)½
(πε₀md³ / 2q²)½
(2q² / πε₀md³)½
Correct(q² / 2πε₀md³)½
Given
Two electrons fixed, separated by 2d, so each is d from the midpoint.
A proton at the midpoint, displaced by x perpendicular to the line, with x ≪ d.
Proton mass m, charge magnitude q.
Asked
The angular frequency of the resulting SHM.
Concept to use
SHM needs a restoring force proportional to the displacement. The proton is attracted to both electrons. When it moves sideways by x, the components along the line cancel by symmetry, and the components pulling it back add. For x ≪ d the distance is essentially d, and the geometry gives a net force −2kq²x/d³ — exactly the F = −kx form.
DiagramAnimatedA restoring force proportional to x is the signature of SHM.
Formula to useF = −2kq²x/d³ = −mω²x → ω² = 2kq²/md³ = q²/(2πε₀md³)
Baby steps
Each electron is at distance r = √(d² + x²) ≈ d from the displaced proton.
Each attraction has magnitude kq²/r²; its component along the displacement direction is that times x/r.
Two electrons, and their sideways components cancel, so the restoring force is 2 × kq²x/r³ → 2kq²x/d³, directed back towards the line.
Compare with F = −mω²x: ω² = 2kq²/(md³).
Substitute k = 1/(4πε₀): ω² = 2q²/(4πε₀md³) = q²/(2πε₀md³).
ω = (q²/2πε₀md³)½.
Answer
ω = (q² / 2πε₀md³)½
Why this option and not the others
Option
Verdict
Reason
(2πε₀md³/q²)½
rule out
The whole expression is upside down. ω must GROW with charge and SHRINK with mass; this does the opposite.
(πε₀md³/2q²)½
rule out
Also inverted, with the factor of 2 misplaced.
(2q²/πε₀md³)½
rule out
Right shape but four times too large inside the root — the 4 in k = 1/4πε₀ was mishandled.
(q²/2πε₀md³)½
keep
ω² = 2kq²/md³ with k = 1/4πε₀.
Shortcut
You never need the derivation to choose here. Ask only which way should ω move? A bigger charge means a stiffer restoring force, so ω must rise with q — q must be on top. A bigger mass means slower oscillation, so m must be underneath. Two of the four options fail on that test immediately, and the remaining pair differ only by a factor of 4 inside the root.
Q 111Left blankCharging by friction · which way electrons move
Assertion: When we rub a glass rod with silk, the rod gets positively charged and the silk gets negatively charged. Reason: On rubbing, electrons from silk cloth move to the glass rod.
Both assertion and reason are true and reason is the correct explanation of assertion.
Both assertion and reason are true but reason is not the correct explanation of assertion.
CorrectAssertion is true but reason is false
Both assertion and reason are false.
Given
A glass rod rubbed with silk.
Asked
The truth of each statement.
Concept to use
Only electrons move in charging by friction — the nuclei stay put. A body becomes positive by LOSING electrons and negative by gaining them. Since the glass ends up positive, electrons must have left the glass and gone to the silk. The reason states the flow in the opposite direction.
DiagramAnimatedElectrons leave the glass; that is why it ends up positive.
Formula to uselose electrons → POSITIVE · gain electrons → NEGATIVE
Baby steps
Assertion. Glass rubbed with silk does end up positive, with the silk negative. True.
Reason. If electrons moved from silk to glass, the glass would gain electrons and become negative. That contradicts the assertion. False.
The correct flow is glass → silk.
So the assertion is true and the reason is false.
Answer
Assertion is true but reason is false
Why this option and not the others
Option
Verdict
Reason
Both true, R explains A
rule out
The reason would make the glass negative, which is the opposite of what the assertion says.
Both true, R does not explain
rule out
The reason is not true at all.
A true, R false
keep
Glass ends positive because it LOSES electrons to the silk.
Both false
rule out
The assertion is a standard textbook result and is correct.
Shortcut
Test the reason against the assertion for consistency before testing it against physics. If the assertion says a body is positive and the reason says it gained electrons, the two are already in conflict, so at most one can be true. That alone eliminates the two “both true” options here — the same structural trick that decides Q105 on this paper.
Q 112Attempted · wrongField of one charge at the other's location
Two point charges Q and −3Q are placed at some distance apart. If the electric field at the location of Q is E⃗, the field at the location of −3Q is:
E⃗
−E⃗
Correct+E⃗/3
Her answer−E⃗/3
Given
Charges Q and −3Q, separated by r.
The field at Q's location (produced by −3Q) is E⃗.
Asked
The field at the location of −3Q.
Concept to use
Each field is produced by the other charge, at the same separation r. So the magnitudes are in the ratio of the charges: Q against 3Q, giving 1 : 3. The directions need care — and the surprise is that they come out the same, not opposite.
DiagramAnimatedBoth fields point the same way along the line.
Formula to useE at Q = k(3Q)/r² ; E at −3Q = kQ/r² → ratio 1/3
Baby steps
Field at Q's location is made by −3Q. A negative charge pulls field lines into itself, so this field points from Q towards −3Q. Call that direction E⃗.
Field at −3Q's location is made by Q. A positive charge pushes field lines away, so this field also points from Q towards −3Q — the same direction.
Magnitude ratio: kQ/r² ÷ k(3Q)/r² = 1/3.
Same direction, one third the size: +E⃗/3.
Answer
+E⃗/3
Why this option and not the others
Option
Verdict
Reason
E⃗
rule out
Would need the two charges to be equal in magnitude. One is three times the other.
−E⃗
rule out
Wrong magnitude and wrong direction.
+E⃗/3
keep
One third the size, and both fields point the same way — from Q towards −3Q.
−E⃗/3
rule out
The magnitude is right but the sign is not. It assumes the two fields oppose, which the drawing shows they do not.
Shortcut
Draw both arrows before deciding the sign. A positive source pushes away, a negative source pulls towards — and when the two charges have opposite signs, those two rules happen to point the same way along the line. Then the magnitude is just the charge ratio, since r is common. Two seconds of drawing replaces all the sign reasoning.
Where it went wrong
The 1/3 was found correctly, so the magnitude reasoning was sound; only the sign came out reversed. The instinct that “opposite charges must give opposite fields” is what misleads here, and it does not hold: the field at each location is made by the other charge, and both happen to point along the line from Q towards −3Q. This is the same direction-of-vector slip flagged in earlier papers — and again the fix is to draw rather than to reason in words.
Q 113Left blankDielectric constant from a force ratio
Two charges placed in air repel each other by a force of 10⁻⁴ N. When oil is introduced between the charges, the force becomes 2.5 × 10⁻⁵ N. The dielectric constant of oil is
2.5
0.25
2.0
Correct4.0
Given
Force in air: F_air = 10⁻⁴ N.
Force in oil: F_oil = 2.5 × 10⁻⁵ N.
Same charges, same separation.
Asked
The dielectric constant K of the oil.
Concept to use
Putting a dielectric between two charges weakens the force by exactly the factor K. So K is simply the ratio of the force in air to the force in the medium — and since the medium force is always smaller, K is always at least 1.
DiagramAnimatedA medium always weakens the force, so K is never below 1.
Formula to useF_medium = F_air / K → K = F_air / F_medium
Baby steps
K = F_air / F_oil = 10⁻⁴ / (2.5 × 10⁻⁵).
Handle the digits and the powers separately: 1/2.5 = 0.4 and 10⁻⁴/10⁻⁵ = 10.
K = 0.4 × 10 = 4.0.
Answer
K = 4.0
Why this option and not the others
Option
Verdict
Reason
2.5
rule out
This is just the digit from the second force, quoted back.
0.25
rule out
The ratio taken upside down. K below 1 would mean the oil strengthened the force, which no dielectric does.
2.0
rule out
Half the correct value; would need F_oil = 5 × 10⁻⁵ N.
4.0
keep
10⁻⁴ divided by 2.5 × 10⁻⁵.
Shortcut
K ≥ 1 always, so divide the bigger force by the smaller one and you cannot get the ratio upside down. That single check kills the 0.25 option before any arithmetic. Vacuum has K = 1 exactly; every real medium is above it.
Q 115Left blankWork done at an angle to the field
There is an electric field in +X direction. If the work done on moving a charge of 0.2 C through a distance of 2 m along a line making an angle of 60° with +X-axis is 4 J. The value of E is
√3 N C⁻¹
4 N C⁻¹
5 N C⁻¹
Correct20 N C⁻¹
Given
Field along +X.
Charge q = 0.2 C moved 2 m at 60° to the field.
Work done W = 4 J.
Asked
The magnitude of E.
Concept to use
Only the part of the displacement along the field does work. So the 2 m at 60° counts as 2 cos 60° = 1 m of useful travel. Everything else in the formula is a straight substitution.
DiagramAnimatedOnly the component of displacement along E does any work.
Formula to useW = qEd cos θ
Baby steps
W = qEd cosθ = 0.2 × E × 2 × cos 60°.
cos 60° = 0.5, so the right-hand side is 0.2 × E × 2 × 0.5 = 0.2E.
Set equal to 4 J: 0.2E = 4.
E = 20 N C⁻¹.
Answer
E = 20 N C⁻¹
Why this option and not the others
Option
Verdict
Reason
√3
rule out
Would arise from using sin or tan of 60° somewhere; neither belongs in W = qEd cosθ.
4
rule out
This is the work in joules, quoted back as a field.
5
rule out
Would follow from forgetting the charge, using W = Ed cosθ instead of qEd cosθ.
20
keep
0.2 × 2 × 0.5 = 0.2, and 4 ÷ 0.2 = 20.
Shortcut
Collect the whole bracket first: qd cosθ = 0.2 × 2 × 0.5 = 0.2. Then E is just W divided by that one number. Doing the multiplication before the division keeps the decimals simple and avoids the usual slip of dropping the charge.
Q 118Left blankWork in a conservative field
An electric field at (x, y) is given by E⃗ = (yî + xĵ) N C⁻¹. The work done (in J) by the electric field in moving a 1 C charge from r⃗ₐ = (2î + 2ĵ) m to r⃗ₒ = (4î + ĵ) m is
Correct0
1
3
2
Given
E⃗ = yî + xĵ.
Charge q = 1 C.
From A(2, 2) to B(4, 1).
Asked
The work done by the field.
Concept to use
This field is conservative, and it comes from the potential V = −xy — check by differentiating: −∂V/∂x = y and −∂V/∂y = x. Once you have V, the work depends only on the two endpoints and not at all on the path taken.
DiagramAnimatedBoth endpoints sit on the same equipotential, xy = 4.
Formula to useE = −∇V with V = −xy ; W = q(Vₐ − Vₒ)
Baby steps
Guess and verify the potential: V = −xy gives −∂V/∂x = y ✓ and −∂V/∂y = x ✓.
Vₐ at (2, 2) = −(2)(2) = −4.
Vₒ at (4, 1) = −(4)(1) = −4.
W = q(Vₐ − Vₒ) = 1 × (−4 − (−4)) = 0.
The two points happen to lie on the same equipotential, xy = 4, so no work is done however you travel between them.
Answer
W = 0 J
Why this option and not the others
Option
Verdict
Reason
0
keep
Both endpoints give V = −4, so they sit on the same equipotential surface xy = 4.
1
rule out
There is no route to this; the potential difference is exactly zero.
3
rule out
Would follow from summing coordinate changes rather than evaluating V.
2
rule out
Likewise not obtainable from V = −xy at these two points.
Shortcut
Spot that xy is the same at both points: 2 × 2 = 4 and 4 × 1 = 4. Since V = −xy, equal xy means equal potential, and equal potential means zero work. That observation takes about five seconds and skips the calculus entirely. Whenever an E-field question gives you two tidy coordinate pairs, test whether some simple product or sum is conserved between them.
Q 122Attempted · wrongNull point between two like charges
Two charges 9 μC and 1 μC are placed at a distance of 30 cm. The position of third charge from 9 μC between them so that it does not experience any force.
7.5 cm
Correct22.5 cm
Her answer5.858 cm
10 cm
Given
Charges 9 μC and 1 μC, 30 cm apart.
A third charge is placed between them.
Distance is measured from the 9 μC charge.
Asked
The distance from the 9 μC charge at which the net force is zero.
Concept to use
For the forces to cancel they must point in opposite directions, which for two like charges happens only between them. The null point sits where the two field magnitudes match — and since field falls as 1/r², the distances go as the square roots of the charges, putting the point farther from the bigger charge.
DiagramAnimatedThe gap splits in the ratio of the SQUARE ROOTS of the charges.
Formula to use9/x² = 1/(30 − x)² → 3/x = 1/(30 − x)
Baby steps
Let the null point be x cm from the 9 μC charge, so 30 − x from the 1 μC.
Equate magnitudes: 9/x² = 1/(30 − x)².
Take the square root of both sides — this is the step that makes it easy: 3/x = 1/(30 − x).
Cross-multiply: 90 − 3x = x, so 4x = 90.
x = 22.5 cm from the 9 μC charge.
Answer
22.5 cm from the 9 μC charge
Why this option and not the others
Option
Verdict
Reason
7.5 cm
rule out
This is the distance from the 1 μC charge, not from the 9 μC. Right point, wrong end measured from.
22.5 cm
keep
The distances split in the ratio 3 : 1, the square roots of 9 : 1.
5.858 cm
rule out
Comes from splitting the gap in the ratio of the charges 9 : 1 rather than of their square roots.
10 cm
rule out
Would be correct only if the charges were in the ratio 4 : 1 by square root, i.e. 4 : 1 in distance.
Shortcut
Square-root the charges and split the gap in that ratio. √9 : √1 = 3 : 1, so the 30 cm divides as 22.5 : 7.5, with the larger share next to the larger charge. No equation needed. And always check which end the question measures from — both numbers appear in the option list.
Where it went wrong
5.858 cm is what you get by splitting in the ratio of the charges (9 : 1) instead of their square roots (3 : 1). The 1/r² in Coulomb's law is what turns the charge ratio into a square-root ratio, and skipping that step is the standard error here. Note also that 7.5 cm is on the option list — the same point measured from the other charge — so the paper is testing the reading as well as the physics.
Q 124Left blankCharge from a measured field
The magnitude of point charge due to which the electric field 30 cm away has the magnitude 2 N C⁻¹ will be
Correct2 × 10⁻¹¹ C
3 × 10⁻¹¹ C
5 × 10⁻¹¹ C
9 × 10⁻¹¹ C
Given
E = 2 N C⁻¹ at r = 30 cm = 0.30 m.
Asked
The magnitude of the point charge.
Concept to use
A direct rearrangement of the point-charge field formula. The only thing to guard is the unit conversion — 30 cm must become 0.30 m before squaring, and squaring makes any slip there large.
DiagramAnimatedE = kq/r², rearranged for the charge.
Formula to useE = kq/r² → q = Er²/k
Baby steps
Convert: r = 30 cm = 0.30 m, so r² = 0.09 m².
q = Er²/k = (2 × 0.09) / (9 × 10⁹).
= 0.18 / (9 × 10⁹) = 0.02 × 10⁻⁹.
q = 2 × 10⁻¹¹ C.
Answer
q = 2 × 10⁻¹¹ C
Why this option and not the others
Option
Verdict
Reason
2 × 10⁻¹¹
keep
Er²/k with r = 0.30 m.
3 × 10⁻¹¹
rule out
Would need E = 3 N C⁻¹ at the same distance.
5 × 10⁻¹¹
rule out
Would need E = 5 N C⁻¹.
9 × 10⁻¹¹
rule out
Uses the 9 from 9 × 10⁹ in the wrong place.
Shortcut
Because k = 9 × 10⁹, the 9 tends to cancel neatly: here 0.18/9 = 0.02 exactly. Watch for that cancellation and the arithmetic collapses to one step. And convert centimetres to metres before squaring, never after — a factor of 100 becomes 10 000 once squared.
Q 130Left blankField from position vectors
A point charge 50 μC is located in the xy-plane at the point of position vector r⃗₀ = (2î + 3ĵ) m. The electric field at the point of position vector r⃗ = (8î − 5ĵ) m is
1200 V m⁻¹
900 V m⁻¹
Correct4500 V m⁻¹
2000 V m⁻¹
Given
q = 50 μC at r⃗₀ = (2, 3).
Field wanted at r⃗ = (8, −5).
Asked
The magnitude of the electric field at that point.
Concept to use
The distance in Coulomb's law is the separation between the two points, not the length of either position vector. So subtract the vectors first, then take the magnitude, then square it. Skipping the subtraction is the trap.
DiagramAnimatedSubtract the position vectors first — a 6-8-10 triangle.
Formula to user⃗_sep = r⃗ − r⃗₀ ; E = kq / |r⃗_sep|²
Its magnitude: √(36 + 64) = √100 = 10 m. A 6-8-10 triangle, which is the paper being kind.
E = kq/r² = (9 × 10⁹ × 50 × 10⁻⁶) / 10².
= (4.5 × 10⁵) / 100 = 4500 V m⁻¹.
Answer
E = 4500 V m⁻¹
Why this option and not the others
Option
Verdict
Reason
1200
rule out
No route to this from a separation of 10 m.
900
rule out
Would need a separation of about 22 m.
4500
keep
Separation 10 m, so E = 9×10⁹ × 5×10⁻⁵ / 100.
2000
rule out
Would need a separation of 15 m.
Shortcut
Look for the Pythagorean triple. Papers almost always arrange the coordinates so the separation is a whole number — 3-4-5, 6-8-10, 5-12-13. If your subtraction gives something untidy, you have probably subtracted in the wrong order or dropped a sign. Here (6, −8) gives exactly 10.
What the twelve have in common
Reading the paper as a whole
Draw the arrows before computing
Three of the five wrong answers — Q99, Q112 and Q122 — would have been caught by
sketching the field arrows at the point in question before touching a formula.
Q
What was done
What the sketch shows
99
Added 9 and 4.5 to get 13.5
Between two LIKE charges the arrows point opposite ways, so they subtract.
112
Chose −E/3
Both fields point from Q towards −3Q — the same way, so +E/3.
122
Split 30 cm in the ratio 9 : 1
Field goes as 1/r², so the split is by √9 : √1 = 3 : 1.
The rule that covers all three: like charges → fields oppose → subtract;
unlike charges → fields reinforce → add. Decide it from the signs first.
Two assertion-reason questions, one structural trick
Q105 and Q111 were both missed, and in both the reason contradicts the assertion
rather than explaining it. When that happens the two cannot both be true, so the two
“both true” options die before any physics is considered.
Q105 — the assertion says both bodies need not be charged; the reason says they
must be oppositely charged. Direct opposites.
Q111 — the assertion says the glass ends positive; the reason says the glass
gained electrons, which would make it negative.
The physics behind both: repulsion is the sure test of charge, not attraction, and
a body turns positive by LOSING electrons.
The blanks were mostly one-liners
Q
The whole method
113
K = F_air / F_medium. One division, and K is never below 1.
115
qd cosθ = 0.2, then E = 4 ÷ 0.2.
124
q = Er²/k, with 30 cm converted to 0.30 m before squaring.
130
Subtract the position vectors: (6, −8) gives 10 m.
118
xy = 4 at both points, so the potentials are equal and W = 0.
Five questions, 20 marks, perhaps four minutes of work between them.
One habit for the next paper
For any question asking “which way”, draw before you calculate. A field
arrow takes two seconds to sketch and cannot be reasoned into pointing the wrong way, whereas a
remembered rule can. This is the fourth paper in a row where a direction or an inequality came
out reversed, and drawing is the one remedy that has not yet been tried systematically.