Charged Metallic Balloon ( understanding a concept)

AI Thread Summary
The discussion revolves around calculating the electric potential on the surface of a metallic balloon with a radius of 34.5 cm and a charge of 1.05 × 10^-8 C. The application of Gauss's law is emphasized to determine the electric field, leading to the formula Voltage = Ed, where d is the distance from the center to the surface. Clarification is provided that the distance d is indeed 34.5 cm, as it represents the radius of the balloon, and the potential is derived from integrating the electric field from infinity to the surface. The concept of electric potential is explained as the work done to bring a unit positive charge from infinity to the balloon's surface.
Bailey
Messages
45
Reaction score
0
the questioon is this:
A balloon of radius 34.5 cm is sprayed with a metallic coating so that the surface is conducting. A charge of 1.05 × 10-08 C is placed on the surface. What is the potential on the balloon's surface?

how i solved it:
-apply the guass law, which gives me the Electrical Field.

-apply the formula : Voltage = Ed ,

where E = electrical field , & d = distance traveled by charges (parrallel to field line)

n it turn out d = 34.5cm. this is where i don't get. why its d = 34.5cm?

its it b/c the positive charge move to the center, which travel a distance of 34.5cm? since if the surface is positively charged, it would repel the " + " charge to the centre & attact the " - " charge to surface...right?
 
Physics news on Phys.org
Bailey said:
how i solved it:
-apply the guass law, which gives me the Electrical Field.
No problem. You should get the field as a function of distance from the center, for points outside of the balloon: E = kq/r^2.
-apply the formula : Voltage = Ed ,
You must integrate from a reference point (usually infinity = 0 potential) to the balloon's surface.

where E = electrical field , & d = distance traveled by charges (parrallel to field line)

n it turn out d = 34.5cm. this is where i don't get. why its d = 34.5cm?
It's not. It turns out that the potential at a point a distance r from the center of the balloon (where r >= the balloon's radius) will equal kq/r.
 
Bailey said:
n it turn out d = 34.5cm. this is where i don't get. why its d = 34.5cm?

its it b/c the positive charge move to the center, which travel a distance of 34.5cm? since if the surface is positively charged, it would repel the " + " charge to the centre & attact the " - " charge to surface...right?
Electrical potential is the work done to bring a unit +charge from infinity to the surface of the sphere, which is:

\int_{\infty}^{34.5} E\cdot ds = kQ(\frac{1}{R}-\frac{1}{\infty}) = \frac{1}{4\pi \epsilon_0}(\frac{Q}{34.5}-0)

AM
 
ah, ic, thanks guy. it seem to clear thing up quite a bit.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged 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