Chapter 23: Q. 7 (page 652)

The irregularly shaped area of charge in FIGURE $Q 23.7$ has surface charge density $η_{i}$ . Each dimension (x and y) of the area is reduced by a factor of $3.163$ .
a. What is the ratio $η_{f} / η_{i}$ , where $η_{f}$ is the final surface charge density?
b. An electron is very far from the area. What is the ratio $F_{f} / F_{i}$ of the electric force on the electron after the area is reduced to the force before the area was reduced?

Short Answer

Expert verified

(a) The ratio of final to initial surface charge density when dimensions (x and y) are reduced by a factor of $3.163 is 10.005 .$

(b) The ratio of the electric force on the electron which is very far from the area after the area of positive charge is reduced to the force before the area was reduced is one.

Step by step solution

Given information (part a)

$I n i t i a l a r e a A_{i} = x x y C h a r g e Q_{i} = Q C h a r g e Q_{f} = Q$

Explanation (part a)

$Surface charge density η = \frac{charge}{area} Ratio of surface charge densities = \frac{η_{f}}{η_{i}} = \frac{\frac{Q_{f}}{A_{f}}}{\frac{Q_{i}}{A_{i}}} \Rightarrow \frac{η_{f}}{η_{i}} = \frac{x \times y}{(\frac{x}{3.163}) \times (\frac{y}{3.163})} \Rightarrow \frac{η_{f}}{η_{i}} = 3.163 \times 3.163 \Rightarrow \frac{η_{f}}{η_{i}} = 10.005$

Given information (part b)

An electron is very far from the area.

Explanation (part b)

Here, the total charge is constant before and after the shrinkage of the area.

Further, as per the given data, the electron is very far from the area and hence it may be taken not only as a point charge. Thus, the force on the electron by the area of positive charge remains constant before and after its shrinkage as the farthest electron shows no impact on the area.