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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

Derive Equation 23.11for the field Edipolein the plane that bisects an electric dipole.

Charge Q is uniformly distributed along a thin, flexible rod of length L. The rod is then bent into the semicircle shown in FIGURE. 23.47

a. Find an expression for the electric field Eat the center of the semicircle.

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Hint: Can you create this charge distribution as a superposition of charge distributions for which you know the electric field?

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