Question

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

Answer 1

a) at the midpoint between 2 rings?
should the E just be added together but using r=10cm?

Answer 2

\[E=F/Q\]

Answer 3

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

Answer 4

(x^2+R^2)^1.3??

Answer 5

oops 1.5

Answer 6

is it between to plates ?

Answer 7

yes

Answer 8

two plates*

Answer 9

yes

Answer 10

\[\epsilon _{0}\ = ?]

Answer 11

8.85*10^-12

Answer 12

k=1/(4pi8.85*10^-12)=8.99*10^9

Answer 13

first convert cm to m

Answer 14

right

Answer 15

.1m diamter, .2m apart

Answer 16

and +40nC=+

Answer 17

+40*10^-9

Answer 18

answer says it should be 0 cuz they cancel each other but i thought it would double since both rings are +

Answer 19

both rings are either + /-

Answer 20

calculate the force, r= d/2

Answer 21

i actually understand now from the set up on cramster except for the part that is raised to 1.5...typo? should be .5

Answer 22

o charge density for ring=Q/(x^2+r^2)^1.5

Answer 23

i mean Q/(2pi*r)

Answer 24

r=d/2

Answer 25

d=0.1m/2 = 0.05m

Answer 26

total net E=E1+E2