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In Problems 63 through 66 you are given the equation(s) used to solve a problem. For each of these

a. Write a realistic problem for which this is the correct equation(s).

b. Finish the solution of the problem

(9.0×109Nm2/C2)(2.0×10-9C)s(0.025m)3=1150N/C

Short Answer

Expert verified

(a) A dipole with charge ±q(2×10-9C)and a distance sbetween them.

At a distance of 0.025mfrom the dipole, the electric field is 1150N/C.

In a dipole, what is the charge separation?

(b) The solution is1mm

Step by step solution

01

Given information and formula used

Given :

(9.0×109Nm2/C2)(2.0×10-9C)s(0.025m)3=1150N/C

Theory used :

Anelectric dipole deals with the separation of the positive and negative charges found in any electromagnetic system.

02

Writing a realistic problem and finding the solution of the problem

(a) Realistic problem :

A dipole with charge ±q(2×10-9C)and a distance sbetween them.

At a distance of 0.025mfrom the dipole, the electric field is 1150N/C.

In a dipole, what is the charge separation?

(b) We can safely assume that r>>s.

1150N/C=(9.0×109Nm2/C2)(2.0×10-9C)s(0.025m)3s=(1150N/C)((0.025m)3(9.0×109Nm2/C2)(2.0×10-9C)=1×10-3m=1mm

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