anonymous
  • anonymous
two 10cm diameter charged rings face each other 20cm apart. both rings are charged to +40nC. what is the electric field strength...
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
a) at the midpoint between 2 rings? should the E just be added together but using r=10cm?
EarthCitizen
  • EarthCitizen
\[E=F/Q\]
anonymous
  • anonymous
found the answer on cramster step by step but not understanding the r: Electric field due to charged ring E = KQx / [ x ^ 2 + R^2 ] ^ 1.5

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More answers

anonymous
  • anonymous
(x^2+R^2)^1.3??
anonymous
  • anonymous
oops 1.5
EarthCitizen
  • EarthCitizen
is it between to plates ?
anonymous
  • anonymous
yes
EarthCitizen
  • EarthCitizen
two plates*
anonymous
  • anonymous
yes
EarthCitizen
  • EarthCitizen
\[\epsilon _{0}\ = ?]
anonymous
  • anonymous
8.85*10^-12
anonymous
  • anonymous
k=1/(4pi8.85*10^-12)=8.99*10^9
EarthCitizen
  • EarthCitizen
first convert cm to m
anonymous
  • anonymous
right
anonymous
  • anonymous
.1m diamter, .2m apart
anonymous
  • anonymous
and +40nC=+
anonymous
  • anonymous
+40*10^-9
anonymous
  • anonymous
answer says it should be 0 cuz they cancel each other but i thought it would double since both rings are +
EarthCitizen
  • EarthCitizen
both rings are either + /-
EarthCitizen
  • EarthCitizen
calculate the force, r= d/2
anonymous
  • anonymous
i actually understand now from the set up on cramster except for the part that is raised to 1.5...typo? should be .5
anonymous
  • anonymous
o charge density for ring=Q/(x^2+r^2)^1.5
anonymous
  • anonymous
i mean Q/(2pi*r)
EarthCitizen
  • EarthCitizen
r=d/2
EarthCitizen
  • EarthCitizen
d=0.1m/2 = 0.05m
EarthCitizen
  • EarthCitizen
total net E=E1+E2

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