Induction in electrostatic equilibrium conductor

AI Thread Summary
In an electrostatically balanced conductor, both the interior and surface remain equipotential. When approaching loaded objects, the surface of the conductor continues to be an equipotential. If it were not, electric field lines could form between regions of excess positive and negative charges. Electric field lines are always perpendicular to the surface of the conductor. Any deviation would cause charge movement until equilibrium is restored.
A13235378
Messages
50
Reaction score
10
Homework Statement
Consider a conductor in electrostatic equilibrium, approach if charged objects so that one region of that conductor has an excess of positive charges and another region with an excess of negative charges.
Relevant Equations
Equipotential regions
We know that both the interior and the surface of an electrostatically balanced conductor are equipotential. My question is if when we approach the loaded objects, the surface of the conductor will continue to be an equipotential. If not, then there could be a field line that left the region with an excess of positive charges going to the region with an excess of negative charges?
 
Physics news on Phys.org
Correct. The electric field lines are perpendicular to the surface. If they were not, charge would move until they are perpendicular.

##\ ##
 
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}...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Back
Top