Question

Two point charges +4 micro coulombs and -10 micro coulombs are placed 10 cm apart form each other in air. A dielectric state of large length and breadth but of thickness 5cm is placed between them.If the relative permittivity of the dielectric is 9, what is the force of attraction between the charges?
a)12N
b)9N
c)5N
d)8N
help me out please..i have to do this question by today..:(

Answer 1

@salini are you viewing this question?

Answer 2

is there any formula for that dielectric(insulator)? i am a beginner here...12th....

Answer 3

me too im in beggining of 12th..but il post the method how i did. i am unable to solve an equation.

Answer 4

it is an insulator?

Answer 5

no it is just a system of charges connected by a wire with negligble capacitance later

Answer 6

before being attached to the wire, the force acting on the charges is F = kq1q2/r^2

Answer 7

then use colomb's law

Answer 8

and after being attached o the capacitor, forces get equally distributed among the two charged bodies. they gain same sign and repel each other with a force f.
the charge on both the charges each = (q1+q2)/2

Answer 9

therefore for the two charges, (q1+q2/2)^2 .

Answer 10

yeah using conservation of charges

Answer 11

f= k[(q1+q2)/2]^2/r^2;
F=-8f (given)
hence, -kq1q2/8r^2 = f [F/-8]
also f = k (q1 + q2/2)^2/r^2 (as shown)
therefore equationg them we get ;
-q1q2/2 = (q1 + q2)^2;
so how do i fing the ratio of q1 and q2 from this equation?

Answer 12

why -8f? wat is that F

Answer 13

oh dude sorry..i was so stupid..i was solving another question..:P

Answer 14

sooooo soo stupid

Answer 15

yeah right can see frm ur display picture how u are!

Answer 16

lol! okay but how do i solve this question?

Answer 17

permitivity of dielectric...this seems new to me i have completed till electric flux....have never seen forces apart from vaccum(that is epsilon knot===peritivity of free space)
will let u know as sooon as i get to know
skool teching or coaching?

Answer 18

coaching

Answer 19

k will get back to this problem....bye!

Answer 20

yeah thanks..:)

Answer 21

hey - i got to know how to do it. we need to find equivalent distance of the force experienced by the charges in the slab in terms of the air medium. then we can find out the force.

Answer 22

?? i think i got some error.

Answer 23

what error?

Answer 24

the force experienced in the air is 9 times the force experienced in slab
http://www.wolframalpha.com/input/?i=9*10^9*4*10*10^-12%2F0.1^2 <-- on air.

Answer 25

the answer is 9 newtons

Answer 26

well, il just tell how i got it -

Answer 27

force along electric slab = K q1q2/9(0.05)^2;
equivalent force in air = Kq1q2/r^2;
therefore Kq1q2/9(0.05)^2 = Kq1q2/r^2;
we get r = 0.15 m
then the force experienced by the charge in air medium totally =
Kq1q2/(0.15 + 0.05)^2;
which we get to be 9N

Answer 28

looks like right ...

Answer 29

yeah thanks..but a friend in my school explained the concept to me..so credit goes to him..:)

Answer 30

problem was quite conceptual ...