Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infinity as your reference point. Compute the gradient of V in each region, and check that it yields the correct field. Sketch V(r ).
Question 2.21: Find the potential inside and outside a uniformly charged so...
- Potential inside a uniformly charged solid sphere
- Potential outside a uniformly charged solid sphere
- Radius of the sphere (R)
- Total charge of the sphere (q)
- Using infinity as the reference point
- Computing the gradient of V in each region
- Checking that the gradient yields the correct field
- Sketching V(r)
Step 1:
To find the potential inside and outside a uniformly charged solid sphere, we need to consider two regions: outside the sphere (r > R) and inside the sphere (r < R).
Step 2:
Outside the sphere (r > R), the electric field is given by E = 1/(4πε0)(q/r^2)(r^), where ε0 is the permittivity of free space, q is the total charge of the sphere, and r is the distance from the center of the sphere.
Step 3:
We can find the potential V(r) outside the sphere by integrating the electric field E with respect to r from infinity (∞) to r. The integral becomes V(r) = -∫[∞ to r] 1/(4πε0)(q/r^2) dr.
Step 4:
Evaluating the integral, we get V(r) = 1/(4πε0)(1/r)(q)|∞ to r = 1/(4πε0)(q/r).
Step 5:
Simplifying the expression, we find V(r) = (q/4πε0)(1/r).
Step 6:
Inside the sphere (r < R), the electric field is given by E = 1/(4πε0)(q/R^3)(rr^), where R is the radius of the sphere.
Step 7:
We can find the potential V(r) inside the sphere by integrating the electric field E with respect to r from infinity (∞) to r and subtracting the integral for the outer region. The integral becomes V(r) = -∫[∞ to r] 1/(4πε0)(1/r^2)(q) dr - ∫[R to r] 1/(4πε0)(1/R^3)(q × r) dr.
Step 8:
Evaluating the integrals, we get V(r) = 1/(4πε0)(q/R) - 1/(4πε0)(q/R^3)((r^2-R^2)/2).
Step 9:
Simplifying the expression, we find V(r) = (q/4πε0)(1/2R)(3 - r^2/R^2).
Step 10:
To compute the gradient of V in each region, we take the partial derivatives of V with respect to the respective coordinates.
Step 11:
Outside the sphere (r > R), the gradient ∇V =1/ (4πε0)(q/r)(∂/∂r) r^ = - 1/(4πε0)(q/r^2) r^.
Step 12:
Simplifying the expression, we find ∇V = - 1/(4πε0)(q/r^2) r^.
Step 13:
Inside the sphere (r < R), the gradient ∇V = 1/(4πε0)(q/2R)(∂/∂r)(3 - r^2/R^2) r^ = -1/ (4πε0)(q/2R)(2r/R^2)(r^) = - 1/(4πε0)(q/R^3)(r) r^.
Step 14:
Simplifying the expression, we find ∇V = - 1/(4πε0)(q/R^3)(r) r^.
Step 15:
Checking that the gradients yield the correct electric field, we see that E = -∇V.
Step 16:
Outside the sphere (r > R), the electric field E = 1/(4πε0)(q/r^2) r^.
Step 17:
Inside the sphere (r < R), the electric field E = 1/(4πε0)(q/R^3)(r) r^.
Step 18:
To sketch V(r), we can plot the potential as a function of the distance r from the center of the sphere. The exact shape of the graph will depend on the values of q, R, and ε0, but we can expect that V(r) will decrease as r increases.
Step 19:
The units of V(r) are q/4πε0R, which can be written as q/4πε0R or q/4πε0×2R depending on the choice of units for charge and permittivity.
Step 20:
This completes the step-by-step explanation of how to find the potential inside and outside a uniformly charged solid sphere, compute the gradient of V in each region, and sketch V(r).
Final Answer
V (r) = −\int_{\infty }^{r}{} E·dl.\begin{cases}& Outside the sphere (r > R) : E =\frac{1}{4\pi \varepsilon _0}\frac{q}{r^2} \hat{r}. \\&Inside the sphere (r < R) : E =\frac{1}{4\pi \varepsilon _0}\frac{q}{R^3}r\hat{r}. \end{cases}
So for r > R: V (r) = −\int_{\infty }^{r}{} \left(\frac{1}{4\pi \varepsilon _0}\frac{q}{\bar{r}^2 } \right)d\bar{r} =\frac{1}{4\pi \varepsilon _0}q\left(\frac{1}{\bar{r} } \right)\mid ^{r}_{\infty }=\frac{q}{4\pi \varepsilon _0}\frac{1}{r} ,
and for r < R: V (r) = −\int_{\infty }^{R}{}\left(\frac{1}{4\pi \varepsilon _0}\frac{q}{\bar{r}^2 } \right)d\bar{r}-\int_{R}^{r}{} \left(\frac{1}{4\pi \varepsilon _0}\frac{q}{R^3}\bar{r} \right)d\bar{r} =\frac{q}{4\pi \varepsilon _0} \left[\frac{1}{R}-\frac{1}{R^3} \left(\frac{r^2-R^2}{2} \right) \right]=\frac{q}{4\pi \varepsilon _0}\frac{1}{2R}\left(3-\frac{r^2}{R^2} \right) .
When r > R, ∇V =\frac{q}{4\pi \varepsilon _0}\frac{\partial}{\partial r}\left(\frac{1}{r} \right)\hat{r} =-\frac{q}{4\pi \varepsilon _0}\frac{1}{r^2}\hat{r}, so E = −∇V = \frac{q}{4\pi \varepsilon _0}\frac{1}{r^2}\hat{r} .
When r > R, ∇V =\frac{q}{4\pi \varepsilon _0}\frac{1}{2R}\frac{\partial}{\partial r}\left(3-\frac{r^2}{R^2} \right) \hat{r}=\frac{q}{4\pi \varepsilon _0}\frac{1}{2R}\left(-\frac{2r}{R^2} \right)\hat{r} =-\frac{q}{4\pi \varepsilon _0} \frac{r}{R^3}\hat{r} ; so E = −∇V = \frac{1}{4\pi \varepsilon _0}\frac{q}{R^3}r\hat{r} .
(In the figure, r is in units of R, and V (r) is in units of q/4\pi \varepsilon _0R.)
Related Answered Questions
The initial configuration consists of a point char...
The theorem is false. For example, suppose the con...
(a) One such point is on the x axis (see diagram) ...
Suppose the n point charges are evenly spaced arou...
E =- \nabla V=-A \frac{\partial}{\partial r...
V=\frac{1}{4 \pi \epsilon_{0}} \int \frac{\...
(a) Potential of +\lambda \text { is } V_...
(a) \nabla^{2} V=-\frac{\rho}{\epsilon_{0}...
(a) E =\frac{1}{4 \pi \epsilon_{0}} \int \...
\rho=\epsilon_{0} \nabla \cdot E =\epsilon...