Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$

= $\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]$

∴NOTE: Graph is attached

8 0

3 years ago

State the law of conservation of energy

storchak [24]

Answer:

In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. ... For instance, chemical energy is converted to kinetic energy when a stick of dynamite explodes.

7 0

2 years ago

Two charges, each q, are separated by a distance r, and exert mutual attractive forces of F on each other. If both charges becom

jeka57 [31]

Answer:

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

we substitute

F = k 4q² / 9 r²

F = k q² r² 4/9

F = ⅔ F₀

3 0

3 years ago

An object releasing energy is evidence of which type of chemical change?

Ray Of Light [21]

Exothermic is the answer to your question

8 0

2 years ago