Class 12 Physics · Original EZT practice
Electric Charges and Fields (Class 12): original worked practice
A correct magnitude can still come with the wrong direction. Track the sign of each charge, convert the units and decide which direction the field or force points before doing the arithmetic.
8 original worked examples, 8 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 17 October 2026:
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Count electrons with the sign attached
Choose the method: For an ordinary body charged by electron transfer, gaining electrons makes the net charge negative; losing them makes it positive.
EZT practice question
Using the stated rounded e, a body has charge −1.12 × 10⁻¹⁸ C. How many excess electrons does that represent?
Show the worked solution
- Divide the charge magnitude by e: (1.12 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) = 7.
- The negative sign means seven more electrons than the neutral body, not seven missing electrons.
Answer: Seven excess electrons.
Check: 7 × (−1.6 × 10⁻¹⁹ C) = −1.12 × 10⁻¹⁸ C under the rounded model.
Watch for this: The calculated charge is approximate because the numerical value of e was rounded.
Now try it yourself
A body has charge +1.44 × 10⁻¹⁸ C under the same approximation. What electron transfer does it represent?
Check your answer
A deficit of nine electrons.
1.44 × 10⁻¹⁸ divided by 1.6 × 10⁻¹⁹ is 9; the positive sign means electrons were lost.
Add charges algebraically
Choose the method: An isolated system conserves total net charge, even when charge moves between its parts.
EZT practice question
Two isolated model bodies begin with +11 nC and −7 nC. Charge redistributes, leaving the first at +1 nC. Find the second charge.
Show the worked solution
- The initial total is +11 − 7 = +4 nC.
- The final total must still be +4 nC, so the second charge is 4 − 1 = +3 nC.
Answer: +3 nC
Check: +1 nC +3 nC equals the initial +4 nC.
Watch for this: The magnitudes 11 and 7 cannot be added as if both charges were positive.
Now try it yourself
The initial charges are −9 nC and +4 nC. The first finishes at −2 nC. Find the second.
Check your answer
−3 nC
The total is −5 nC, so −2 nC plus the second charge must equal −5 nC.
Convert separation before squaring
Choose the method: For fixed point charges in vacuum, the force magnitude is k|q₁q₂|/r². Decide attraction or repulsion separately.
EZT practice question
Point charges +4 μC and −6 μC are 40 cm apart in vacuum. Estimate the force magnitude and direction.
Show the worked solution
- Use q₁ = 4 × 10⁻⁶ C, q₂ = −6 × 10⁻⁶ C and r = 0.40 m.
- F = (9.0 × 10⁹)(24 × 10⁻¹²)/(0.40)² = 0.216/0.16 = 1.35 N.
- The signs differ, so each charge is pulled toward the other.
Answer: Approximately 1.35 N, attractive.
Check: N m²/C² × C²/m² leaves newtons. The force acts along the line joining the charges.
Watch for this: Using 40 instead of 0.40 changes r² by a factor of 10,000.
Now try it yourself
Estimate the force between +5 μC and +8 μC separated by 60 cm in vacuum.
Check your answer
Approximately 1.0 N, repulsive.
The magnitude is 9.0 × 10⁹ × 40 × 10⁻¹² /0.36 = 1.0 N; the charges have the same sign.
Compare before calculating again
Choose the method: With the charges fixed, the force varies as 1/r². Scale the separation ratio and square it.
EZT practice question
Two point charges exert a force of magnitude F at separation r. What is the magnitude at 3r/2?
Show the worked solution
- The distance becomes 1.5 times its original value.
- The force multiplier is 1/(1.5)² = 4/9.
Answer: 4F/9
Check: A larger separation must give a smaller force for the same charges.
Watch for this: Multiplying by 2/3 only accounts for one power of distance.
Now try it yourself
If the separation becomes 2r/5, what is the force magnitude?
Check your answer
25F/4
The reciprocal square multiplier is (5/2)² = 25/4.
Keep field direction separate from test-charge force
Choose the method: A source field points in the direction of force on a positive test charge. F = qE reverses direction for a negative charge.
EZT practice question
Find the field 30 cm from an isolated +2 μC point charge in vacuum, then the force on a −3 nC test charge there. Neglect its effect on the source.
Show the worked solution
- E = kQ/r² = (9.0 × 10⁹)(2 × 10⁻⁶)/0.09 = 2.0 × 10⁵ N/C, away from the source.
- The force magnitude is 3 × 10⁻⁹ × 2.0 × 10⁵ = 6.0 × 10⁻⁴ N.
- The negative test charge feels force opposite the field, toward the source.
Answer: E ≈ 2.0 × 10⁵ N/C outward; force ≈ 6.0 × 10⁻⁴ N inward.
Check: The field unit N/C times charge C gives force N.
Watch for this: A negative test charge does not reverse the source field itself.
Now try it yourself
A uniform field points east with magnitude 4.0 × 10⁴ N/C. Find the force on +7 nC.
Check your answer
2.8 × 10⁻⁴ N east.
A positive charge feels force along the field: 7 × 10⁻⁹ × 4.0 × 10⁴ N.
Add field vectors, not charge magnitudes
Choose the method: At a point away from the charges, add each field with its direction. A zero resultant has no defined direction.
EZT practice question
Equal positive point charges are fixed at x = −a and x = +a, where a > 0. What is their combined field at the origin?
Show the worked solution
- Their distances from the origin are equal, so their field magnitudes there are equal.
- The left charge gives a rightward field; the right charge gives a leftward field. These cancel.
Answer: Zero field at the origin; its direction is undefined.
Check: A small test charge placed exactly there would have zero net electric force in this two-source model.
Watch for this: A zero vector does not have a unique direction. Nor does zero field at one point mean both source charges vanish.
Now try it yourself
Replace the right charge with an equal negative charge. What happens to the two field directions at the origin?
Check your answer
Both point right, from the positive source toward the negative source, so the magnitudes add.
The field points away from a positive charge and toward a negative charge.
Use the full separation in p = qd
Choose the method: A dipole has equal opposite point charges. Its moment points from negative to positive. Its far field has a 1/r³ scale at a fixed angle.
EZT practice question
A dipole has charges ±3 μC separated by 4 mm. Find its moment. Then compare its field at distances r and 3r along the same ray, with r much larger than the separation.
Show the worked solution
- p = qd = (3 × 10⁻⁶)(4 × 10⁻³) = 1.2 × 10⁻⁸ C m.
- The direction is along the line from the negative charge to the positive charge.
- For the far-field comparison at the same angle, E(3r)/E(r) ≈ (1/3)³ = 1/27.
Answer: p = 1.2 × 10⁻⁸ C m, from − to +; far-field magnitude is approximately 1/27 at 3r.
Check: The dipole can have zero total charge and a nonzero dipole moment because the charges are separated.
Watch for this: The 1/r³ approximation is not a general near-field formula and does not remove angular dependence.
Now try it yourself
A dipole with charges ±4 μC has separation 7 mm. Find p.
Check your answer
2.8 × 10⁻⁸ C m, from the negative charge to the positive charge.
Multiply 4 × 10⁻⁶ C by the full 7 × 10⁻³ m separation.
Measure the angle from the chosen normal
Choose the method: For a flat surface in a uniform field, signed flux is Φ = EA cos θ, with θ measured between E and the area normal.
EZT practice question
A 20 cm by 15 cm rectangle is in a uniform field of 3.0 × 10⁴ N/C. Its chosen normal makes 60° with the field. Find the signed flux.
Show the worked solution
- Convert the area: A = 0.20 × 0.15 = 0.030 m².
- Φ = (3.0 × 10⁴)(0.030) cos 60° = 900 × 0.5.
Answer: +450 N m²/C
Check: With the chosen normal reversed, θ becomes 120° and the flux becomes −450 N m²/C.
Watch for this: The angle between the field and the plane is not θ in this formula. Flux is not a count of physical particles moving through the surface.
Now try it yourself
For the same rectangle and field, what is the flux if the chosen normal makes 90° with the field?
Check your answer
Zero.
The field lies parallel to the plane and cos 90° = 0.
Try a mixed set
Choose a method or explanation before opening each answer.
1. In the rounded e model, is a net charge of +3.5e possible for an ordinary isolated body charged only by electron transfer?
Show answer and reason
No.
Electron transfer changes the charge by an integer multiple of e, which may be positive, negative or zero.
2. An artist draws twice as many field lines for the same charge diagram. Has the physical field doubled?
Show answer and reason
No.
The number drawn is a convention. Relative line density within a consistent drawing represents field strength.
3. Do two charged conducting spheres always obey the point-charge formula exactly when r is measured between their centres?
Show answer and reason
No.
Nearby conductors can polarise and redistribute charge. The point-charge approximation needs a suitable size-to-separation limit; exact spherical-symmetry results need their stated conditions.
4. Why can two distinct electrostatic field lines not cross at a point with nonzero field?
Show answer and reason
A crossing would assign different tangent directions to the same nonzero field vector.
There is one resultant field vector at the point. This argument does not assign a direction to a zero vector.
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Questions about this practice page
- Is this the whole chapter?
- No. This is selected original practice. The linked chapter also covers Gauss law and its applications, which are outside this set.
- Which constants do these questions use?
- Numerical estimates use k = 9.0 × 10⁹ N m²/C² and e ≈ 1.6 × 10⁻¹⁹ C. BIPM defines the exact elementary charge as 1.602176634 × 10⁻¹⁹ C.
- Is charge always a positive whole-number multiple of e?
- No. The signed integer n in Q = ne can be negative, zero or positive for the ordinary electron-transfer model here.
- Does flux have a sign?
- Yes. It depends on the chosen normal. For a closed surface, the standard convention uses the outward normal.
- Are these copied from the NCERT Electric Charges and Fields exercises?
- No. These are original selected questions with checks, not an official exercise index or a complete answer key.