D.2 · IB Physics
Electric field of point charges
Each arrow shows the direction of the force on a positive test charge. Drag the charges, change their strength, or switch to a potential map or field lines. The field is always the vector sum, the potential the scalar sum.
Predict first
Start from the default dipole with Field arrows and the test charge ticked. Double the charge size from 2.0 nC to 4.0 nC. What happens to the field at the test charge?
E = kQ ÷ r² is directly proportional to Q at fixed distances, so doubling every charge doubles the field. The direction is unchanged.
Check it in the model: Move the Charge size, |Q| slider from 2.0 to 4.0 nC and compare the Field at test charge readout.
positive charge negative charge. Canvas width = 1.60 m. Drag a charge or the small test dot.
V = kQ ÷ r (scalar sum)
F = qE
k = 8.99×10⁹ N m² C⁻²
Try this
- Choose the dipole with field arrows and drag the test charge along the line midway between the charges. In which direction does the field point there, and why is it never zero?
- Choose "Two + arrows" and place the test charge on the line between the charges. Find the point where E = 0, then move it off the line and see what happens to the arrow.
- Switch to the potential map for the dipole. Compare the value of V at the midpoint with the field E there, and explain why one is zero and the other is not.
Common error: field is a vector and potential is a scalar. Add fields with components, but add potentials as signed numbers, so equal and opposite charges give V = 0 at the midpoint while E there is not zero.