Surface charge density and distance?

Question

Hi,

The surface charge density on an infinite charged plane is -2.2 x 10^-6 C/m2. A proton is shot straight away from the plane at 1.9 x 10^6 m/s. How far does the proton travel before reaching its turning point?

Any idea on how to start this one?


Thanks!

Thank you! That is the right answer. I was wondering if the V in V=-Ed is velocity. Where does the V=-Ed equation come from?

Thanks!

Answer 1

There is a formula for the electric field of in infinitely charged plane. This can be evaluated from Gauss's Law for electric fields.
First of all, you have to notice that the electric field lines are always perpendicular to the plane. Then, making a Gaussian surfice like a prism with area A and height H cutting the plane, we have:

2*E*A = sigma*A/ep0 => E = sigma/(2*ep0),

where sigma is the surface charge density and ep0 is the permitivity of vacuumt. We see that the field is homogenous (like the field of gravity near Earth), so the potential is:

V = -E*d,

where d is the distance from the plane

Now, from the law of conservation of energy for the proton, we have:

m*v^2/2 = -e*E*d

d = - m*v^2/(2*e*E)

d = -ep0*m*v^2/(e*sigma)

Calculating, you get:

d = 0.15 m

where we have used:

ep0 = 8.85 x 10^(-12) C^2/(N*m^2)
m = 1u = 1.66 x 10^(-27) kg
e = 1.6 x 10^(-19) C

Answer 2

I’ve got an idea, but I don’t know proton’s charge and mass, so let them be e Coulombs and m kg respectively. According to Coulomb’s law the elementary force of attraction dF=(c*ds)*e*r/|r|^3, where given c=2.2·10^(-6) C/m^2, ds is elementary area on infinite plane, vector r = x*i +y*j +z*k is connecting ds and the proton, point [0,0,0] being at proton gun; thus F= e·c·∫ r·ds/|r|^3 for plane area = 0 until infinity in 360° around;
Here let x=p*cos(t), y=p*sin(t), while |r|^2 = xx+yy+zz = p^2 +z^2 and ds=(p*dt)*dp;
F= e·c·∫ r·ds/|r|^3 = e·c·∫ (p*cos(t)*i +p*sin(t)*j +z*k)*(p·dp·dt) /|r|^3;
Now, integration {for t=0 until 2pi} results in i and j components being compensated, while k component is present and F = 2pi·e·c·z ∫ p·dp/|r|^3;
F= 2pi·e·c·z ∫ p·dp / (z^2+p^2)^1.5 {for p=0 until infinity};
F= -2╥·e·c·z / sqrt(z^2+p^2) {for p=0 until infinity} = 2╥·e·c = constant;
Now energy balance: F*z = 0.5*m*v^2; hence z = m*v^2 / (4╥·e·c);

Answer 3

hi asf12312, the container via -? the container that too uniform container will be appealing so directed in direction of the plate on both area of the plate. That uniform container = ? / 2 epsilonnot So the gap from 6 ?C on effective coordinate will be x in this variety of fashion (a million/4pi epsilonnot) * q/x^2 = ? /2 epsilonnot Cancelling epsilonnot on both area, x^2 = (a million/2 pi) * q/? So x^2 = 0.3184 So x = 0.564 m for this reason at a distance of 0.564 m correct from the Q on +ve x axis the Enet = 0 yet on adverse x axis there won't be able to be this variety of degree the position E internet = 0 because the gap at which the container will develop into equivalent is properly interior of 0.6 m. notwithstanding the forces might want to no longer be cancelled as both act interior an same direction. yet on the decrease back area of the plate the fields are opposite to at least one yet another yet not in any respect develop into equivalent in value