Help with Electric Field on a Square Insulating Sheet?

AI Thread Summary
To estimate the electric field above a square insulating sheet with uniform charge Q, the key is to recognize that when the distance r_2 is much greater than the length L of the sheet, the sheet can be approximated as a point charge. The electric field for a point charge is given by E = Q/(4*pi*epsilon_0*r_2^2). This simplification allows for easier calculations, as the orientation of the sheet does not affect the electric field at large distances. Understanding this approximation is crucial for solving the problem effectively. The discussion highlights the importance of recognizing charge distribution behavior at varying distances.
isabellef
Messages
10
Reaction score
0

Homework Statement



A square insulating sheet whose sides have length L is held horizontally. The sheet has a charge of Q spread uniformly over its area.

a) Estimate the electric field at a point located a distance r_2 above the center of the sheet. Use the appropriate approximations based on the fact that r_2 >> L.

Homework Equations



E * dA = Q_encl/epsilon_0

The Attempt at a Solution



The area of the square sheet is L^2, which can go into the equation above. From that, I know that the electric field for r_2 << L is E = Q /(2*epsilon_0*L^2), but I can't seem to figure out the field for r_2 >> L.

Any suggestions?
 
Physics news on Phys.org
From far away a sheet of paper looks like a spot or a point. The approximation I think they want is to treat the sheet of charge as a point charge? You got it from there?
 
Not quite. Do you mind explaining a little further?
 
If you are far from a sheet of charge it does not matter what the orientation of the sheet of charge is the electric field still points towards you. Far from a distribution of charge the charge distribution can be approximated as a point charge.
 
If you have any object that has a uniform charge density; as your sheet does, your object can be treated as a point particle if the radius is MUCH MUCH greater than the overall size of the initial object. That's what the r_2>>>L is telling you. You should know how to calculate the E-field for a point particle.
 
Oh okay, now I understand. Thank you for the help!
 
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}...
Back
Top