Adding up the push of many charges

Science · Electricity & Magnetism · ages 18-19

Name ______________________ Date ____________

The electric field of an isolated positive point charge points away from it, and its magnitude is proportional to 1/r², where r is the distance from the charge.
Circle one: True False
How is the electric field at a point defined?
- Force on a tiny positive test charge divided by that charge
- Force on a tiny positive test charge multiplied by that charge
- Total source charge divided by the distance to the test point
A uniformly charged line has density λ = 4 nC/m. How much charge is on a segment 0.25 m long?
- 4 nC
- 1 nC
- 16 nC
At a point on the axis of a uniformly charged ring, why do the sideways field components cancel?
- Opposite arcs give equal sideways components in opposite directions
- Each arc gives an axial field with no sideways component of its own
- The test charge removes the sideways components before fields add
A thin ring of radius R has uniform positive line density λ. At distance z > 0 on its axis, which integral gives E_z? θ is the angle around the ring in radians, and k is Coulomb's constant.
- E_z = ∫₀π kλRz/(R²+z²)^(3/2) dθ
- E_z = ∫₀²π 2kλRz/(R²+z²)^(3/2) dθ
- E_z = ∫₀²π kλRz/(R²+z²)^(3/2) dθ
For a finite rod with uniform positive charge, an integral predicts that field magnitude grows as you move much farther away than the rod's length. What does the far-field check tell you?
- Accept it: a finite positive rod produces a field that grows at large distances
- Recheck it: a finite charged rod should approach an inverse-square far field
- Ignore it: a finite rod has no point-charge limit at distances beyond its length
You divide a charged sheet into small area patches dA to calculate its field. Which density gives each patch's charge through dq = density × dA?
- Surface density: charge per unit area
- Linear density: charge per unit edge length
- Volume density: charge per unit occupied volume
A student adds the magnitudes of two electric field contributions that point in different directions. What must change in the calculation?
- The fields cannot be added at all, because electric fields never obey superposition
- The fields can be added as magnitudes, even when their directions are different
- The fields must be added as vectors, keeping each contribution's direction

Answer key

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Adding up the push of many charges W1-mt_IIcVqWapC1-s1

True · A positive source repels a positive test charge. Its field magnitude is kQ/r².
Force on a tiny positive test charge divided by that charge · E = F/q, where F is the force on a tiny positive test charge q at the point.
1 nC · For constant line density, segment charge is density times length: 4 × 0.25 = 1 nC.
Opposite arcs give equal sideways components in opposite directions · Equal opposite arcs are equally far from the axial point. Their sideways components are equal and opposite, while their axial components add.
E_z = ∫₀²π kλRz/(R²+z²)^(3/2) dθ · Use dq = λR dθ and the axial projection z/√(R²+z²). Integrate once around the full ring, from 0 to 2π.
Recheck it: a finite charged rod should approach an inverse-square far field · The rod is finite and has positive total charge Q. Far beyond its length, its leading field magnitude is kQ/r², which decreases with distance.
Surface density: charge per unit area · An area patch uses surface density σ, measured in C/m². Multiplying by dA gives dq = σ dA.
The fields must be added as vectors, keeping each contribution's direction · Electric fields obey vector superposition. Add corresponding signed components; magnitudes alone discard the directions needed for cancellation.