Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

1.) To one decimal place, what is the best approximation to the electric field m

ID: 1645557 • Letter: 1

Question

1.) To one decimal place, what is the best approximation to the electric field magnitude in V/cm at point g?

2.) To one decimal place, what is the best approximation to the electric field magnitude in V/cm at point h?

3.) To one decimal place, what is the best approximation to the electric field magnitude in V/cm at point i?

4.) To one decimal place, what is the best approximation to the electric field magnitude in V/cm at point j?

5.)To one decimal place, what is the best approximation to the electric field magnitude in V/cm at point k?

6.) To one decimal place, what is the best approximation to the electric field magnitude in V/cm at point l?

7.) What would happen to a positively charged liquid poured at point i? Why?

8.) What would happen to a positively charged liquid poured at point c? Why?

5 points 5 points Save Answer Question 10 betaoen and nzne diagram The figure and tablc below are usnd in Questions 10-22 The figur shaws the equipatentials for cach intrger waltage betaen and indding-4 Vand +4 V for en errangement of charges as 9hONn Aasume that-4 V ie the minimum potenral ancl +4 8 lhe maximum lor ary point within he diagram The labie gives the closest dstance from a point to neighboring equpoleni al, either the one rat one soit over or re one rat is one von higher. htor en erangament od harnge rem p 1 eighbern usd e o that 4 Cm 25

Explanation / Answer

We know that E = -dV/dr
so for approximate calculations
electric field at a point can be written as
E = -del(V)/del(r) where del(V) is the potential drop in the vicinity of the point in distance del(r)

a. at point g
del(V) = 2V
del(r) = 1.7 + 0.65
E = 2/2.35 = 0.851 V/cm
b. at point h
its inside a conductor with constant potential, so, E = 0 V/cm

c. at point i
its inside a conductor with constant potential, so, E = 0 V/cm

d. at point j
del(V) = 2V
del(r) = 0.6 + 0.5
E = 2/1.1 = 1.818 V/cm

e. at point k
del(V) = 2V
del(r) = 0.6 + 1.5
E = 2/2.1 = 0.952 V/cm

f. at point l
del(V) = 2V
del(r) = 1.5 + 0.6
E = 2/2.1 = 0.952 V/cm