Question

charge Q is distributed uniformly along a semicircle of radius a. which formula below gives correct magnitude of electric field at centre of circle?

Answer 1

See my attachment in:
http://openstudy.com/users/goutham1995#/updates/4fae5db2e4b059b524f9a07e

Answer 2

\[1)E = kQ/(pi) a \]
\[2. E = kQ/pi a ^{2}\]
\[3. E= K * 2Q/pi*a\]
4.E=k*2Q/pie * a^2
5.E=k*2Q/a^2

Answer 3

these are the options

Answer 4

Can you find which ones to rule out because of lack of homogeneity?

Answer 5

yeah good idea!

Answer 6

but it will take a long time

Answer 7

How? electric field is Coulomb's constant times charge over distance squared

Answer 8

yeah but we have to do that for all the options

Answer 9

It only takes a few seconds to see that 1) and 3) cannot be expressions for an electric field.

Answer 10

yeah..because if 1 is not possible, then 3 too is not possible since it only has a constant along with it

Answer 11

other three are matching

Answer 12

But look at the denominator!

Answer 13

There MUST be a distance squared in the denominator. If not, no chance to end up with an electric field

Answer 14

yeah i understood that..hence option 2 , 4 and 5 are possible

Answer 15

Yes. Now, 2) is not possible either because fields add up as vectors, not as scalars.

Answer 16

i didnt understand that ..:(

Answer 17

????????

Answer 18

The method to derive the field is here: study each of the steps. They have numbers so it makes it easier to say later which needs elaboration.

Answer 19

yeah understood the derivation. but i didnt understand how you eliminated option 2

Answer 20

As all charges are at the same distance a of centre, then the answer given looks like a sum of magnitudes. It is not so important.

Answer 21

oh it looks like k*lambda

Answer 22

So do you know which answer is good now?

Answer 23

option d or e

Answer 24

I mean, you've just said you understand the derivation; so you must use the result to find which is the right expression.

Answer 25

The result is in box 4. How does it apply to your problem?

Answer 26

\[2*K*\lambda/a [\sin]\left(\begin{matrix}90 \\ 0\end{matrix}\right)\]

Answer 27

Yes, so the final part is just sin90°-sin0° = 1
and you have to work out λ in terms of Q and a.

Answer 28

okay thanks