Electric field inside a sphere of charge

AI Thread Summary
To find the electric field inside a charged sphere, Gauss's Law is applied by considering a smaller sphere within the charged sphere. The charges outside this inner sphere do not contribute to the electric field because their effects cancel each other out, resulting in a net electric field of zero. This cancellation occurs as the electric field vectors from external charges point in various directions, leading to a vector sum of zero. The total flux through the inner sphere confirms this, reinforcing the concept that only enclosed charge influences the electric field inside. Understanding this principle is crucial for grasping electrostatics in spherical charge distributions.
maccha
Messages
49
Reaction score
0
So I understand that to find the electric field inside a sphere of charge you can use Gauss's Law by drawing a sphere inside, enclosing the charge. But, intuitively, I don't understand it.. why don't the charges outside this sphere contribute to the electric field? I know the total flux through the sphere you've drawn will be zero, but I don't know, it just doesn't make sense visually to me.
 
Physics news on Phys.org
The fields due to the charges outside the sphere all point in different directions. The vector sum of these fields, as it turns out, adds up to zero.
 
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...
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