You are actually adding two pieces: (1) change in potential from ∞ to R, (2) change in potential from R to r.gracy said:I don't know why are we subtracting these two?
Just becauseDoc Al said:You are actually adding two pieces:
Yep!gracy said:Negative sign!
And why adding them gives potential at r<R?Doc Al said:You are actually adding two pieces
The work done against the electric field (which is the potential) in going from "A" to "C" equals the work done from A to B plus the work done from B to C.gracy said:And why adding them gives potential at r<R?
Right.gracy said:potential at infinity is zero,as we move in the direction of electric field the potential increases .Moving from infinity to R gives potential at R (integration with upper and lower limits i.e final and initial positions being infinity and R),right?
Number 2: The change in potential. (The potential itself is defined as work from infinity to r.)gracy said:Then Moving from R to r (initial and final points being R and r respectively)gives
1)potential at r
OR
2)Change in potential while going from R to r
which one?
The potential due to a uniformly charged sphere can be expressed as V = kQ/R, where k is the Coulomb's constant, Q is the charge of the sphere, and R is the distance from the center of the sphere.
The potential due to a uniformly charged sphere is directly proportional to the electric field. This means that as the potential increases, the electric field also increases.
Yes, the potential due to a uniformly charged sphere can be negative. This occurs when the charge of the sphere is negative. A negative potential means that work needs to be done to move a positive test charge from infinity to a point near the sphere.
The potential due to a uniformly charged sphere decreases with distance from the center of the sphere. This can be seen in the expression V = kQ/R, where R is in the denominator. As R increases, V decreases.
No, the potential due to a uniformly charged sphere is not affected by the shape of the sphere. As long as the charge is uniformly distributed, the potential will be the same regardless of the shape of the sphere.