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anonymous
 4 years ago
two 10cm diameter charged rings face each other 20cm apart. both rings are charged to +40nC. what is the electric field strength...
anonymous
 4 years ago
two 10cm diameter charged rings face each other 20cm apart. both rings are charged to +40nC. what is the electric field strength...

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a) at the midpoint between 2 rings? should the E just be added together but using r=10cm?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0found 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

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.0is it between to plates ?

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.0\[\epsilon _{0}\ = ?]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0k=1/(4pi8.85*10^12)=8.99*10^9

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.0first convert cm to m

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0.1m diamter, .2m apart

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0answer says it should be 0 cuz they cancel each other but i thought it would double since both rings are +

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.0both rings are either + /

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.0calculate the force, r= d/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i actually understand now from the set up on cramster except for the part that is raised to 1.5...typo? should be .5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0o charge density for ring=Q/(x^2+r^2)^1.5
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