A ring that has radius a lies in the z 0 plane with its center at...

Question

Question

Answer 1

Answer #1

answer) we know the formula

Ez=kQz/z²+a²)^3/2

a) Ez(0.1,a)=kQ*0.1a/((0.1a)²+(a²))^3/2=0.099kQ/a²

b)Ez(0.4a,a)=kQ0.4a/((0.4a²+a²)^3/2=0.377kQ/a²

c)Ez(0.7,a)=kQ0.7a/((0.7a)²+a²)^3/2=0.385kQ/a²

d) Ez(a)=kQa/(a²+a²)^3/2=0.354kQ/a²

e)Ez(1.6a)=kQ1.6a/(1.6a)²+a²)^3/2=0.238kQ/a²

f) the figure is below 2 2 43210 1.234- O 凸

Answer 2

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