Problem statement

A positive point charge of 50 µC sits in the xy plane at the point with position vector r0 = 2i + 3j. Find the electric field vector E, and its size, at the point with position vector r = 8i − 5j. All lengths are in metres.

We assume the charge is a point, and it sits alone in vacuum. This means no other charge or material changes the field, and we can use Coulomb's law directly.

We assume everything is static. This means the charge does not move and no magnetic effects appear.

We assume k = 1/(4π ε0) = 8.988 × 10^9 N m²/C², where ε0 = 8.854 × 10^-12 C²/(N m²) is the permittivity of free space (a constant that sets the strength of electric forces).

Symbols: q is the source charge. The vector r0 is the arrow from the origin to the charge, and r is the arrow from the origin to the field point. The letters i and j are unit vectors (arrows of length 1) along the x and y axes. E is the electric field: the force per unit of positive charge.

The big idea in plain words

A charge fills the space around it with a field. The field at a point tells us what push a small positive test charge would feel there. Coulomb's law gives that push. We divide by the test charge to get the field. The push points along the straight line from the source to the point. So we need the size, from the distance, and the direction, from the arrow between the two points.

The solution in each frame

Lab frame

Everything is at rest, so the lab frame is the only frame we need. The full solution is in the next section.

Centre-of-mass frame

This frame does not apply. There is only one charge, and it is fixed, so there is no motion about a centre of mass.

Medium frame

This frame does not apply. The charge is in vacuum, so there is no medium to move with.

The solution step by step

First we find the arrow from the charge to the field point, because the field points along it. The arrow from the charge at r0 to the point at r is r − r0. We subtract the components one at a time.

Next we find the distance R between the two points, because Coulomb's law needs it. The two components of r − r0 are the two short sides of a right triangle, and R is the long side. Pythagoras's theorem gives its length.

Now we find the size of the field, because we know R. Coulomb's law says two point charges q and qt push each other with the force F = k q qt / R². The field is the force per unit test charge, so we divide by qt.

Last we write the direction, because the question asks for the vector. The unit vector (an arrow of length 1) along r − r0 is (r − r0)/R. For a positive charge, E points along it. We multiply the size by this unit vector.

The field is about kV/m, at an angle of degrees from the x axis. A 1 µC test charge placed there would feel a force of N.

Five methods

1. Force method

The free-body diagram is for a small positive test charge qt placed at the point. The only force on it is the push F from the source, along the line from the source. Coulomb's law gives F = k q qt / R². Dividing by qt gives E = k q / R² = kV/m, along (r − r0)/R.

2. Energy method

The potential (energy per unit charge) of a point charge at distance R is V = k q / R. The field points in the direction in which the potential falls fastest, and its size is the slope. So the size of E is the drop in V per metre, E = −dV/dR.

3. Graph method

The third graph plots V against R. Draw the tangent line at your R. Its slope is dV/dR, and the size of E is minus that slope. The first graph plots |E| against R on log scales and is a straight line of slope −2, because E ∝ 1/R². Read |E| at your R.

4. Calculus method

We take the potential V(x, y) = k q / sqrt((x − x0)² + (y − y0)²) and use E = −grad V, which means Ex = −∂V/∂x and Ey = −∂V/∂y. The symbol ∂ means we change only x (or only y) and hold the other fixed. We use the chain rule: for V = k q R^-1, ∂V/∂x = −k q R^-2 × ∂R/∂x, and ∂R/∂x = (x − x0)/R.

5. Relative-position method

The key is the arrow r − r0 between the source and the point. Here its parts are (, ) m. The unit vector is (, ). For the book values the parts are 6 and −8, a 3-4-5 triangle scaled by 2, so R = 10 m and the unit vector is (0.6, −0.8) with no calculator.

Fastest: the relative-position method, which reads the unit vector straight off the displacement. Most reliable: the force method with the vector formula, because it is Coulomb's law with nothing hidden.

What each slider changes

Parameters

Exam traps

• Using the position vector r instead of r − r0. The distance and direction come from the arrow between the charge and the point, not from the origin.
• Subtracting in the wrong order. For a positive charge E points from the charge to the point, so use r − r0, not r0 − r.
• Forgetting to divide by R again in the vector form. The vector is k q (r − r0) / R³, with R cubed, because (r − r0) already has length R.
• Mixing units. µC must become C (× 10^-6) and V/m can become kV/m (÷ 1000).
• Giving only the size. The question asks for the vector and its size, so give both.