Potential Difference and Potential Near a Charged Sheet

AI Thread Summary
The discussion focuses on calculating the potential difference between two points near a charged sheet with surface charge density +σ. The electric field generated by the sheet is constant at E = σ/(2ε₀) and directed away from the sheet. For Part A, the potential difference V_AB is derived as V_AB = -E(y_1 - y_2), indicating a linear relationship based on the vertical positions of points A and B. In Part B, when considering the potential at a distance y_1 from the sheet, the potential is evaluated to be infinite, as the reference point at y = ±∞ is set to zero. The calculations confirm that the approach taken is correct.
Badger
Messages
28
Reaction score
0

Homework Statement


Let {\rm A} = \left(x_1,y_1 \right) and {\rm B} = \left( x_2,y_2 \right) be two points near and on the same side of a charged sheet with surface charge density + \sigma . The electric field \vec{E} due to such a charged sheet has magnitude E = \frac {\sigma}{2 \epsilon_0} everywhere, and the field points away from the sheet, as shown in the diagram. View Figure
184996.jpg


Part A
What is the potential difference V_{\rm AB} = V_{\rm A} - V_{\rm B} between points A and B?

Part B
If the potential at y = \pm \infty is taken to be zero, what is the value of the potential at a point V_A at some positive distance y_1 from the surface of the sheet?
choices are:
1. infinity
2. negative infinity
3. 0
4. -E * y_1


Homework Equations


\int_{\rm B}^{\rm A} \vec{C} \cdot d\vec{\ell} = \int_{x_2}^{x_1} C_x\,dx + \int_{y_2}^{y_1} C_y\,dy<br /> = C_x (x_1 - x_2) + C_y(y_1 - y_2)

V_{\rm AB}= -\int _B^A \vec{E}\cdot d\vec{l}



The Attempt at a Solution


Part A.
V_{\rm AB} = V_{\rm A} - V_{\rm B}= \left(-E\right)\left(y_{1}-y_{2}\right)

Part B.
I figure I'd use the equation I got in part A and set the bottom of the E field at y=0.

In which case
V = -E (y_1 - infinity) = infinity

am i on the right track?
 
Physics news on Phys.org
Looks right to me.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top