NEB Class 11 • Physics • Worked Practice

Electric Field Practice Set: NEB 11 Physics

Work from direction diagrams to inverse-square magnitude, component addition and field-line interpretation. Mark model selection and physical checks, not only final numbers.

Concept, calculation and diagram rounds
Worked superposition and zero-field examples
Conductor questions and timed challenge

The field direction is defined for a positive test charge.

Source check

Keep every problem tied to the field definition

The CDC secondary curriculum and CDC Physics Grade 11 page provide Nepal context. Review the Electric Field guide and study plan. The worked model follows OpenStax E=F/q and point-source field.

Round 1 • Concepts

Answer before calculating

Define electric field and its SI unit.
Why is a positive test charge used for direction?
Does inserting a larger test charge create a larger source field?
What is the difference between E and F?
Why can two fields cancel although both sources remain?
Why do field lines not cross?
What does line density represent?
State conductor properties at electrostatic equilibrium.

Answer framework

Name source charges, chosen point and positive-probe convention. State E as a property of the source configuration at that point; force becomes qE after a charge is placed. Field lines are a drawing convention, not material paths.

Round 2 • Point fields

Worked single-source problems

1. Field from +Q

Q=+4.0 μC at r=0.30 m gives E=(8.99×10⁹)(4.0×10⁻⁶)/0.090≈4.00×10⁵ N/C, radially away.

2. Field from −Q

Q=−6.0 nC at r=0.12 m gives E≈(8.99×10⁹)(6.0×10⁻⁹)/0.0144≈3.75×10³ N/C, toward the source.

3. Force from field

A +2.0 nC charge in 5.0×10⁴ N/C south experiences F=1.0×10⁻⁴ N south. A −2.0 nC charge experiences the same magnitude north.

4. Reverse for source charge

E=9.0×10⁴ N/C at r=0.20 m. |Q|=Er²/k≈(9.0×10⁴)(0.040)/(8.99×10⁹)≈4.0×10⁻⁷ C. Direction reveals the sign.

Round 3 • Collinear sources

Draw arrows before comparing magnitudes

Equal like charges at midpoint

Two +3 μC charges are 0.40 m apart. At the midpoint each gives E≈(8.99×10⁹)(3×10⁻⁶)/(0.20)²≈6.74×10⁵ N/C in opposite directions, so resultant is zero.

Equal opposite charges at midpoint

Replace the right charge by −3 μC. Both midpoint fields point from positive toward negative, so E≈1.35×10⁶ N/C toward the negative charge.

Zero field for unequal like charges

+9Q and +Q are separated by d. Between them, 9/x²=1/(d−x)² gives 3/x=1/(d−x), so x=3d/4 from +9Q, or d/4 from +Q.

For opposite charges, the between-region fields point the same way, so cancellation must be sought outside, on the side of the smaller-magnitude source. Direction reasoning eliminates impossible algebraic roots.

Round 4 • Two dimensions

Resolve fields on common axes

Right-angle sources

At P, one source produces 3.0×10⁴ N/C east and another 4.0×10⁴ N/C north. Resultant is 5.0×10⁴ N/C at 53.1° north of east.

Equal charges on a perpendicular bisector

+2 μC charges at (±0.30,0) m and P=(0,0.40) m. Each distance is 0.50 m and each field is about 7.19×10⁴ N/C. Horizontal components cancel; each vertical component is 0.8E, so total is about 1.15×10⁵ N/C upward.

Dipole on the perpendicular bisector

If the right source becomes negative, vertical components cancel and horizontal components add toward the negative charge. Reuse geometry but redraw directions; do not reuse component signs.

Round 5 • Diagrams and conductors

Six sketches with oral explanations

Single positive charge.
Single negative charge.
Equal dipole.
Equal like pair.
Unequal opposite pair.
Parallel plates with edge effects noted.

For each, state start/end rules, tangent direction, density meaning and why lines do not cross. Then explain why an ideal conductor has zero internal field in electrostatic equilibrium and why the field just outside is normal to its surface.

Extension: sketch a hollow conducting shell with no internal charge. Explain shielding through charge rearrangement rather than saying the metal “blocks electricity.”

Marking grid

Diagnose the first error

Evidence	One mark
Field point and sources labelled	Geometry
Away/toward arrows correct	Direction
SI distances and charges	Units
Each k\|Q\|/r² magnitude	Model
Components added	Vector
N/C and physical statement	Interpretation

Retest the first failed row with different charge signs after two days. Keep geometry constant once, then rotate the entire configuration to test whether reasoning is coordinate-independent.

Timed challenge

Forty-minute mixed set

Find E 15 cm from −8 nC.
Find force on +5 nC and −5 nC at that point.
Compare field after source doubles and distance triples.
Solve midpoint fields for +Q/+Q and +Q/−Q.
Locate zero field for +4Q and +Q separated by 60 cm.
Solve the perpendicular-bisector field of two equal charges.
Sketch a dipole and unequal like pair.
Explain conductor equilibrium and shielding.
Critique crossed field lines in a student drawing.

Key checks: task 3 changes by 2/9. Task 5 lies 40 cm from +4Q and 20 cm from +Q. For every task, a negative placed charge reverses force but not the source field.

Use PhET Charges and Fields only after committing to arrows. For online or physical NEB tuition, call 9846662070.

Frequently asked questions

Questions about the electric-field practice set

Should I calculate magnitude before direction?

No. Draw away/toward arrows first so later component signs have a physical basis.

Where can zero field occur for like charges?

Between them, closer to the smaller charge when magnitudes differ. Solve only after verifying opposite field directions.

Where can zero field occur for opposite charges?

Not between them, because fields point the same way there; if it exists, seek outside on the smaller-charge side.

How should I mark field-line diagrams?

Check origins and ends, arrows, tangent direction, relative density, proportional line count and absence of crossings.

What score shows readiness?

Aim for 80% with no repeated geometry, direction, unit, vector or interpretation error on a fresh set.

Where can I get help with field problems?

Call 9846662070 for current online or physical NEB tuition options and fees.

References and next steps

Sources and related study guides

Review Electric Field Study Guide and the Electric Charges guide after marking. Curriculum scope and sources were checked on 2 August 2026; follow current CDC, NEB and college instructions if requirements change.

Extension design

Construct a field with a required direction

Place two equal positive charges symmetrically on the x-axis. Any point on the positive y-axis has cancelling horizontal components and an upward resultant. To create a rightward resultant instead, place an equal positive–negative pair on the x-axis and choose a point on its perpendicular bisector; both horizontal components point toward the negative charge.

Reverse problem

At P, a source produces 2.0×10⁵ N/C east. A second source must make the resultant 2.0×10⁵ N/C north. Its required field is the vector difference: −2.0×10⁵ east plus 2.0×10⁵ north, magnitude 2.83×10⁵ N/C at 135° from east. Source sign and location must then be chosen to generate that direction at P.

Reverse problems force you to treat E as a vector design quantity rather than a number attached to a charge.

Graph and data round

Extract charge from an inverse-square test

A student measures E at several distances from an isolated spherical source and plots E vertically against 1/r² horizontally. In the point-charge region, gradient equals k|Q|. If the fitted gradient is 1.80×10⁴ N·m²/C, then |Q|≈(1.80×10⁴)/(8.99×10⁹)≈2.00 μC. Vector direction identifies sign.

A nonzero intercept may signal background field or sensor offset. Curvature at small r can show finite source size, probe disturbance or distance-reference error. Scatter should be handled with repeat readings and a fit, not by choosing only points that match theory.

Scaling checkpoint

At 0.10 m the field is E₀. At 0.25 m it should be E₀(0.10/0.25)²=0.16E₀ for an unchanged point source. Compare this ratio before calculating kQ.

This round joins mathematical linearisation with physical model checking and explains why a single field reading cannot test an inverse-square law.

Final challenge: design two different source arrangements that produce the same electric field vector at P. Specify positions and charges, verify every contribution, and explain why matching E at one point does not make the fields identical everywhere. Then place a negative test charge at P and state its force without recalculating the source field. This tests superposition, non-uniqueness and the E–F distinction together.

Mark a final seven-point solution grid: geometry, directions, individual distances, SI conversion, field magnitudes, vector sum and physical interpretation. A calculator result earns no direction mark. Exchange one reverse problem with a classmate and require them to draw the required resultant before selecting source signs or locations. After marking, change one source sign and solve again without copying component signs.

Finally, verify every complete result independently and carefully with symmetry or a limiting-distance argument before accepting it as physically plausible.

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