Surface Charge Density and electric field

Thread starter PatrickGeddes
Start date Apr 21, 2012
Tags

Charge Charge density Density Electric Electric field Field Surface Surface charge density

AI Thread Summary

The discussion focuses on calculating the electric field produced by a nonconducting wall with a surface charge density of 8.55 µC/cm² at a distance of 8.55 cm from the wall. The relevant equations include the surface charge density formula and the electric field formula, E = σ/2ε₀. A participant expresses uncertainty about whether the distance from the wall affects the electric field calculation, suggesting a need for clarification on the influence of distance in this context. The consensus indicates that for distances small compared to the wall's dimensions, the electric field can be approximated without significant adjustments for distance. The conversation emphasizes understanding the relationship between surface charge density and electric field strength.

Apr 21, 2012

PatrickGeddes

Homework Statement

A nonconducting wall carries charge with a uniform density of 8.55 µC/cm2.
(a) What is the electric field 8.55 cm in front of the wall if 8.55 cm is small compared with the dimensions of the wall?

Homework Equations

σ= Q/A
E=σ/2Eo

The Attempt at a Solution

E=8.55/2 x 8.854x10^-12

Physics news on Phys.org

Apr 21, 2012

PatrickGeddes

Don't think this is right, I feel like the factor that the 8.55cm from the wall should make a difference. Is that right?

Thread 'I've come across another fluid pressure problem I don't understand'

Neglect gravity other than that of water; neglect friction. F1=ρgsh=1000*10*1*(1+4)=50000 N; F1=ρgsh=1000*10*0.1*4=4000 N; F3=50000-4000=46000 N?

View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

View full post »

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

View full post »