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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

2.0×1012m/s2=(1.60×10-19C2)E(1.67×10-27kg)E=Q(8.85×10-12C2/Nm2)(0.020m)2Q

Short Answer

Expert verified

(a) What is the strength of the electric field between two parallel conducting planes when a proton is discharged in an area between planes with an acceleration of 2×1012m/s2?

(b) The solution is 7.36×10-11C

Step by step solution

01

Given information and formula used

Given :

2.0×1012m/s2=(1.60×10-19C2)E(1.67×10-27kg)E=Q(8.85×10-12C2/Nm2)(0.020m)2Q

Theory used :

The electric field between two conducting parallel plates is the potential difference divided by the distance by which they are separated.

02

Writing a realistic problem and finding the solution of the problem 

(a) Realistic Problem :

What is the strength of the electric field between two parallel conducting planes when a proton is discharged in an area between planes with an acceleration of 2×1012m/s2?

(b) Solution :

2.0×1012m/s2=(1.60×10-19C2)E(1.67×10-27kg)E=(2.0×1012m/s2)(1.67×10-27kg)(1.60×10-19C2)=2.08×104N/C

Also,

E=Q(8.85×10-12C2/Nm2)(0.020m)2Q=(2.08×104N/C)(8.85×10-12C2/Nm2)(0.020m)2=7.36×10-11C

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Most popular questions from this chapter

Air "breaks down" when the electric field strength reaches 3.0×106N/c, causing a spark. A parallel-plate capacitor is made from two 4.0cm×4.0cm electrodes. How many electrons must be transferred from one electrode to the other to create a spark between the electrodes?

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