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

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A spherically symmetric charge distribution with a net positive charge Q affects the electric field direction based on the distance from its center. For r < R, the electric field E(r) is directed radially inward, despite the positive charge, due to the nature of the potential function provided. For r > R, the electric field is directed radially outward, consistent with the behavior of electric fields from positive charges. The confusion arises from interpreting the potential function, which indicates the electric field's behavior within the sphere. Understanding the implications of charge distribution is crucial for accurately determining electric field direction.
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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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