Question

Why must the test charge \({\rm{q}}\) in the definition of the electric field be vanishingly small?

Short answer

The test charge \({\rm{q}}\) must be vanishingly small so that it does not affect the electric field of the other charge.

Step-by-step solution

Electric field

Electric field is a vector quantity defined as the force acting per unit charge. Mathematically,

$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{E}}\mathbf{=}\frac{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}}{\mathbf{q}}$

Here,$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$ is the force acting on the test charge $\mathit{q}$.

The SI unit of electric field is newton per coulomb $\mathbf{(}\mathit{N}\mathbf{/}\mathit{C}\mathbf{)}$.

The test charge must be vanishingly small

The test charge must be vanishingly small to ensure that the test charge does not alter the electric field that the other charge produces \(\vec E\).