Charges Moving in Uniform Fields
- Apply projectile-motion methods to a charge in a uniform electric field
- Use energy conservation to find the speed of an accelerated charge
- Compare the electric force with gravity for a charged particle
It is a projectile problem with a different constant
A charge in a uniform field feels a constant force F = qE, so it undergoes constant acceleration a = qE/m. Every method from Physics 1 kinematics applies unchanged: the motion perpendicular to the field is constant-velocity, the motion along it is constant-acceleration, and time is the shared variable. A charge entering a parallel-plate region sideways traces a parabola, for exactly the same reason a thrown ball does.
Choosing the method
If the problem states a potential difference and asks for a speed, use energy: the work qΔV becomes kinetic energy, and no geometry is needed. If it gives plate length and separation and asks how far the particle deflects or whether it escapes, use kinematics — horizontal motion sets the time in the field, and the vertical equations give the deflection. Choosing wrongly is not fatal but doubles the work.
Is gravity negligible?
For an electron or proton in a laboratory field, the electric force is many orders of magnitude larger than the weight, so gravity is routinely ignored. For a charged oil drop or dust particle it is not: the Millikan experiment works precisely by balancing the two. The test is to compute both qE and mg and compare. A question that supplies a mass in kilograms rather than as a particle name is usually signaling that gravity matters.
An electron is accelerated from rest through a potential difference of 250 V. Find its final speed. (m = 9.11 × 10⁻³¹ kg, e = 1.60 × 10⁻¹⁹ C)
- 1.Work done on the charge: W = qΔV = (1.60 × 10⁻¹⁹)(250) = 4.00 × 10⁻¹⁷ J.
- 2.All of it becomes kinetic energy: ½mv² = 4.00 × 10⁻¹⁷ J.
- 3.v² = 2(4.00 × 10⁻¹⁷)/(9.11 × 10⁻³¹) = 8.78 × 10¹³.
- 4.v = √(8.78 × 10¹³) ≈ 9.4 × 10⁶ m/s.
The acceleration is qE/m, not qE. Forgetting to divide by the mass gives an answer in newtons where meters per second squared is wanted, and the unit check catches it immediately.
A proton and an electron are each accelerated from rest through the same potential difference. Compared with the proton, the electron ends up with:
A charged particle enters a uniform field perpendicular to the field lines. Its path in the field region is:
The acceleration of a charge q of mass m in a uniform field E is:
Read whether the question asks for speed, time or deflection before choosing a method. Speed from a voltage is one line with energy; deflection from geometry needs kinematics, and attempting either with the other tool wastes most of the time allowed.
Answer the 3 checkpoints as you read.
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