How Does Charge Distribution Affect Electric Field Direction in a Sphere?

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SUMMARY

The discussion focuses on the electric field direction and magnitude resulting from a spherically symmetric charge distribution with a net positive charge Q within a radius R. The electric potential V is defined as V(r) = (kQ/R)(-2 + 3(r^2/R^2)) for r < R and V(r) = (kQ/R) for r > R. It is established that for r < R, the electric field E(r) is directed radially inward, while for r > R, it is directed radially outward. This conclusion arises from the interpretation of the electric potential function and the nature of electric fields around charged distributions.

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Homework Statement



A spherically symmetric charge distribution has net positive charge Q distributed within a radius of R.
Its electric potential V as a function of the distance r from the center of the sphere is given by the following.
V(r)=\frac{kQ}{R}( -2+3{\frac{r^2}{R^2}})for r<R
V(r)=\frac{kQ}{R} for r>R
https://www.physicsforums.com/file:///page5image8000 https://www.physicsforums.com/file:///page5image8160 https://www.physicsforums.com/file:///page5image9232 Express all algebraic answers in terms of the given quantities and fundamental constants.

  1. (a) For the following regions, indicate the direction of the electric field E(r) and derive an expression for its magnitude.

    i. r < R____ Radially inward ____ Radially outward

    ii. r > R____ Radially inward ____ Radially outward

    The answer to (i), i.e., r<R is radially inward.

Homework Equations

The Attempt at a Solution


How can the field due to a positive charge be radially inward?
For (ii), it's radially outward, which is fairly straightforward, because field lines will originate radially from the sphere, but inside, INWARD??
 
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The problem states that the distribution has a NET positive charge. It doesn't say that it is positive throughout. Look at the given potential function...
 

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