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The irregularly shaped area of charge in FIGURE Q23.7 has surface charge densityηi. Each dimension (x and y) of the area is reduced by a factor of 3.163.

a. What is the ratio ηf/ηi , where ηfis the final surface charge density?

b. An electron is very far from the area. What is the ratioFf/Fi 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 3.163is10.005.

(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

01

Given information (part a)

InitialareaAi=xxyChargeQi=QChargeQf=Q

02

Explanation (part a)

Surface charge densityη=chargeareaRatio of surface charge densities=ηfηi=QfAfQiAiηfηi=x×yx3.163×y3.163ηfηi=3.163×3.163ηfηi=10.005

03

Given information (part b)

An electron is very far from the area.

04

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.

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