Question

A very small object with mass 6.90 × 10^−9 kg and positive charge 8.30 × 10^−9 C is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density 5.90 x 10^-8 C/m^2. The object is initially 0.530 m from the sheet.

What initial speed must the object have in order for its closest distance of approach to the sheet to be 0.170 m ?

Answer 1

help guys?

Answer 2

I'm new to Gauss' law, so this will probably be incorrect.

Answer 3

\[\phi =\frac{\sigma A}{\epsilon_0}=\int\limits_{}^{}E(x)dA\]

Answer 4

E is not a function of x, so This simplifies to:\[\frac{\sigma A}{\epsilon_0}=E \int\limits_{}^{} dA=EA\]

Answer 5

Sorry, that should have been a half in front of all the fluxes (multiply all epsilon 0s by 2), as only half of the flux goes through the top.

Answer 6

\[E=\frac{\sigma}{2 \epsilon_0}\]

Answer 7

\[F=Eq\]

Answer 8

Okay, let's assume that's true, and the rest is easy.

Answer 9

\[F=\frac{\sigma q}{2\epsilon_0}=ma\]

Answer 10

\[6.9*10^{-9}a=\frac{(8.3*10^{-9})(5.9*10^{-8})}{2(8.85*10^{-12})}\]

Answer 11

Then I use this to solve for a?

Answer 12

\[a \approx 4*10^3 \pm0.006\]

Answer 13

Do you understand what I've done so far?

Answer 14

Anyway, so we have\[a=-4*10^3\]\[s(0)=0.53\]
At \[s=0.17\]
We want\[v=0\]

Answer 15

I've got my co-ordinate system muddled, make that a positive acceleration.

Answer 16

\[v^2=u^2+2as\]Are you familiar with this formula?

Answer 17

\[0=u^2+2*(4*10^3)*(-0.36)\]

Answer 18

-0.36 is the change in position, displacement.

Answer 19

\[u=\sqrt{8*10^3*0.36 }\approx55.67m/s\]

Answer 20

Let's check this. \[v=u+at=-56+4*10^3t=0\]\[t=0.014\]

Answer 21

\[s=ut+0.5at^2=-56(0.014)+0.5(4*10^3)(0.014)^2=-0.392\]

Answer 22

Which is close enough to the real -0.36 for comfort.

Answer 23

so, is that -.392 the initial speed or 55.67?

Answer 24

Can someone help me out?

Answer 25

why would you ask that?

Answer 26

You could have indicated when you weren't getting it initially. It's the latter.