Chapter 21.1, Problem 16SSC

To determine

To Explain: The method of detection of a field at a point, determination of magnitude of the field, and how to choose the magnitude of the test charge.

Explanation of Solution

Introduction:

Electric field is a physical quantity that gives relationship between the force exerted on a test charge and the size of a test charge. It is a vector quantity. If $F_{onq}$ is the electrostatic force on the test charge $q$ at some point, the electric field $\left(E\right)$ on that charge at the same point is given by,

  $E=\frac{F_{onq}}{q}$

This expression clearly indicates that the electric field on a charge is independent of its size.

To detect the field at a point − If the test charge placed at that point experiences a force then one can say that there exist a field at that point, because any charge experiences a force of attraction or repulsion from the electric field.

To determine the magnitude of the filed $\left(E\right)$ − The ratio of the magnitude of force $\left(F\right)$ on the test charge to the magnitude of test charge $\left(C\right)$ gives the magnitude of the field. That is,

  $\text{Electric field strengh = }\frac{\text{Force}}{\text{Charge}}$

  $E=\frac{F}{C}$ .

To choose the magnitude of the test charge − The test charge must be chosen in such a way that its magnitude must be very small compared to that of the charges producing the field. Because the test charge also interacts with other charges and this interaction should be small for precise measurement.