Physics 2Electric Force, Field and PotentialHS-PS3-5

Fields of Several Charges and Field Lines

Add the field vectors from each charge; field lines show the direction, and their spacing shows the strength.

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With several charges, each one makes its own field, and the field at a point is their vector sum, for the same reason that forces add. Field lines then let you see the whole pattern at a glance.

Why it works

The force on a test charge is the vector sum of the forces from each charge. Divide every force by the test charge and the same sum holds for fields: $\vec E = \vec E_1 + \vec E_2 + \cdots$. Field lines are drawn so that at every point they run along $\vec E$, and so that they crowd together where $E$ is strong. Since the field has only one direction at each point, two lines can never cross.

Key idea

Add the field vectors from each charge. Field lines start on positive charges and end on negative ones, point along $\vec E$, are closer where the field is stronger, and never cross. They show the direction of the force, not the path a charge must follow.
Two equal positive charges: the lines bend apart, and the field is zero at the midpoint.
The same idea as arrows: the field of a positive and a negative charge, drawn on a grid.

Worked examples

Worked example: The middle of a dipole

A $+2.0\,\text{nC}$ and a $-2.0\,\text{nC}$ charge are $0.20\,\text{m}$ apart. Find the field at the midpoint.
  1. Each charge is $0.10\,\text{m}$ away: $E_1 = E_2 = (8.99 \times 10^{9})\dfrac{2.0 \times 10^{-9}}{(0.10)^2} = 1798\,\text{N/C}$.
  2. The positive charge's field points away from it, toward the negative charge; the negative charge's field points toward itself. Same direction.
  3. $E = 1798 + 1798 = 3.60 \times 10^{3}\,\text{N/C}$, toward the negative charge.

Answer: $3.60 \times 10^{3}\,\text{N/C}$

Worked example: Fields at a right angle

A $+2.0\,\text{nC}$ charge is at the origin and a $+4.0\,\text{nC}$ charge at $(0.30\,\text{m}, 0.40\,\text{m})$. Find the field at P $= (0.30\,\text{m}, 0)$.
  1. From the origin, $0.30\,\text{m}$ away: $E_1 = (8.99 \times 10^{9})\dfrac{2.0 \times 10^{-9}}{(0.30)^2} = 200\,\text{N/C}$, in the $+x$ direction.
  2. From the charge $0.40\,\text{m}$ above P: $E_2 = (8.99 \times 10^{9})\dfrac{4.0 \times 10^{-9}}{(0.40)^2} = 225\,\text{N/C}$, in the $-y$ direction.
  3. $E = \sqrt{200^2 + 225^2} = 301\,\text{N/C}$.

Answer: $301\,\text{N/C}$

Common mistake

Thinking field lines are the paths charges must follow, or that two lines can cross. A field line shows the direction of the force, which sets the direction of the acceleration, not of the velocity; a moving charge can cut across lines, just as a thrown ball does not follow the downward pull of gravity. And lines never cross, because at a crossing the field would have two directions at once.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.

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Sources and further reading

  • OpenStax, College Physics 2e, section 18.5 Electric Field Lines: Multiple Charges
  • OpenStax, University Physics Volume 2, section 5.6 Electric Field Lines
  • Next Generation Science Standards, HS-PS3-5
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