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.

For teachers: free to print, copy and use in class. Every new worksheet has new values and its own answer key. A link to ExamProblems on your course page is appreciated.

Name: ____________________   Date: ____________

1.Easy
++4 µC−−5 µCr = ?

Point charges of +4 µC and −5 µC exert a force of 2 N on each other. How far apart r are they? Give the answer in cm. Use k = 8.99 × 10⁹ N·m²/C².

q1 = +4 µCq2 = −5 µCF = 2 N
r = cm
2.Easy
+q₁+q₂r

Two point charges repel each other with a force of 6 N. The first charge is halved and the distance between the charges is tripled. Find the new force F′.

F = 6 Nq₁′ = 0.5·q₁r′ = 3·r
F′ = N
3.Medium
++1 µC++4 µCE = 040 cm

Charges of +1 µC and +4 µC are 40 cm apart. At what distance x from the +1 µC charge, on the line between them, is the electric field zero?

q1 = +1 µCq2 = +4 µCd = 40 cm
x = cm
4.Medium
++8 µC−−5 µCP60 cm

A charge of +8 µC and a charge of −5 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 = +8 µCq2 = −5 µCr1 = 45 cmr2 = 15 cm
E = N/C
5.Hard
5 cm10 cm+q1​ = +5 µC+q3​ = +6 µC+q2​ = +4 µC

Three positive point charges sit at the corners of a right triangle: q₃ = 6 µC is at the right-angle corner, q₁ = 5 µC is 5 cm from it along one side and q₂ = 4 µC is 10 cm from it along the other side. Find the magnitude of the net electric force F on q₃.

q1 = 5 µCq2 = 4 µCq3 = 6 µCr13 = 5 cmr23 = 10 cm
F = N
6.Hard
+ + + + + +− − − − − −e⁻1.5 × 10⁷ m/s8 cmE = 2500 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 8 cm long and the field between them is 2500 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 = 8 cmE = 2500 N/Ce = 1.602 × 10⁻¹⁹ Cme = 9.109 × 10⁻³¹ kg
y = mm

Answer key

  1. 1. r = 29.98 cm
  2. 2. F′ = 0.3333 N
  3. 3. x = 13.33 cm
  4. 4. E = 2.353 × 10⁶ N/C
  5. 5. F = 110 N
  6. 6. y = 6.253 mm

Full step-by-step solutions: practise this topic on examproblems.com.