Coulomb’s law & electric fields worksheet with answers

Free printable Coulomb’s law & electric fields worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
++1 µC−−6 µCr = 10 cm

Two point charges of +1 µC and −6 µC are 10 cm apart. Find the magnitude of the electric force F between them. Use k = 8.99 × 10⁹ N·m²/C².

q1 = +1 µCq2 = −6 µCr = 10 cm
F = N
2.Easy
+q₁+q₂r

Two point charges repel each other with a force of 34 N. The first charge is made four times as large and the distance between the charges is halved. Find the new force F′.

F = 34 Nq₁′ = 4·q₁r′ = 0.5·r
F′ = N
3.Medium
++10 µC−−10 µCP60 cm

A charge of +10 µC and a charge of −10 µC are 60 cm apart. Point P lies on the line between them, 45 cm from the first charge. Find the magnitude of the electric field E at P.

q1 = +10 µCq2 = −10 µCr1 = 45 cmr2 = 15 cm
E = N/C
4.Medium
++3 µC−−3 µCP50 cm

A charge of +3 µC and a charge of −3 µC are 50 cm apart. Point P lies on the line between them, 37.5 cm from the first charge. Find the magnitude of the electric field E at P.

q1 = +3 µCq2 = −3 µCr1 = 37.5 cmr2 = 12.5 cm
E = N/C
5.Hard
+ + + + + +− − − − − −e⁻1.5 × 10⁷ m/s4 cmE = 1400 N/C

An electron enters the uniform field between two parallel plates moving parallel to them at 1.5 × 10⁷ m/s. The plates are 4 cm long and the field between them is 1400 N/C. Electron charge 1.602 × 10⁻¹⁹ C, mass 9.109 × 10⁻³¹ kg; ignore gravity. How far y is it deflected sideways when it leaves the plates? Give the answer in mm.

v₀ = 1.5 × 10⁷ m/sL = 4 cmE = 1400 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm
6.Hard
+ + + + + +− − − − − −e⁻5 × 10⁷ m/s10 cmE = 4200 N/C

An electron enters the uniform field between two parallel plates moving parallel to them at 5 × 10⁷ m/s. The plates are 10 cm long and the field between them is 4200 N/C. Electron charge 1.602 × 10⁻¹⁹ C, mass 9.109 × 10⁻³¹ kg; ignore gravity. How far y is it deflected sideways when it leaves the plates? Give the answer in mm.

v₀ = 5 × 10⁷ m/sL = 10 cmE = 4200 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm

Answer key

  1. 1. F = 5.394 N
  2. 2. F′ = 544 N
  3. 3. E = 4.44 × 10⁶ N/C
  4. 4. E = 1.918 × 10⁶ N/C
  5. 5. y = 0.8754 mm
  6. 6. y = 1.477 mm

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