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If you already know the electric field, you can find the change in electrostatic potential between two points A and B by integrating the field along an arbitrary path joining these two points, VB – VA = − ³* Ë - d5. Recall from Gauss's law that the magnitude of the electric field E(r) at a radial distance r ≤ R from the center of a uniformly charged sphere with total charge Q and radius R is given by E(r)= while the field at r > Ris 1 Q. 47€ R³ 1Q 4π€or² E(r)= If the charge Q is positive, the field points radially out. We will now analyze the electric potential due to a uniform spherical charge distribution. (a) Draw a qualitatively correct graph of the electric field strength as a function of r, E(r), due to a uniform spherical charge distribution. Technically, we should keep track of the direction of E

If you already know the electric field, you can find the change in electrostatic potential between two points A and B by integrating the field along an arbitrary path joining these two points, VB – VA = − ³* Ë - d5. Recall from Gauss's law that the magnitude of the electric field E(r) at a radial distance r ≤ R from the center of a uniformly charged sphere with total charge Q and radius R is given by E(r)= while the field at r > Ris 1 Q. 47€ R³ 1Q 4π€or² E(r)= If the charge Q is positive, the field points radially out. We will now analyze the electric potential due to a uniform spherical charge distribution. (a) Draw a qualitatively correct graph of the electric field strength as a function of r, E(r), due to a uniform spherical charge distribution. Technically, we should keep track of the direction of E

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Problem 3 If you already know the electric field, you can find the change in electrostatic potential between two points A and B by integrating the field along an arbitrary path joining these two points, B VB - V₁ = -√³Ē. A while the field at r > Ris Recall from Gauss's law that the magnitude of the electric field E(r) at a radial distance r < R from the center of a uniformly charged sphere with total charge Q and radius R is given by E(r): = Ē.ds. 1Q 47€ R³ E(r): = 1 Q 4π€ ²* If the charge Q is positive, the field points radially out. We will now analyze the electric potential due to a uniform spherical charge distribution. (a) Draw a qualitatively correct graph of the electric field strength as a function of r, E(r), due to a uniform spherical charge distribution. Technically, we should keep track of the direction of E
Transcribed Image Text:Problem 3 If you already know the electric field, you can find the change in electrostatic potential between two points A and B by integrating the field along an arbitrary path joining these two points, B VB - V₁ = -√³Ē. A while the field at r > Ris Recall from Gauss's law that the magnitude of the electric field E(r) at a radial distance r < R from the center of a uniformly charged sphere with total charge Q and radius R is given by E(r): = Ē.ds. 1Q 47€ R³ E(r): = 1 Q 4π€ ²* If the charge Q is positive, the field points radially out. We will now analyze the electric potential due to a uniform spherical charge distribution. (a) Draw a qualitatively correct graph of the electric field strength as a function of r, E(r), due to a uniform spherical charge distribution. Technically, we should keep track of the direction of E
by using positive values for field pointing out and negative for field pointing in, but since the field all points out everywhere, we don't have to worry about that. Taking the electric potential at the sphere's surface (at r = R) to be equal to zero (ie setting our starting point at the surface,) draw a qualitatively correct graph of V(r), the potential as a function of radial distance from the center. Note that we DO need to be careful about the sign of V! The idea in this part is to draw your graph using the information in the E graph, but NOT do any calculations. Check that your V graph and E graph are consistent with each other at r = 0, at r < R, at r = R, and at r > R. Also check that your V graph is consistent with our choice of V=0 at r=R. (b) What should the sign of AV be for r > R? Should V increase or decrease with r (i.e. if we start at r = R and move out, will V go down or up, or does it do one and then the other? How did you decide? What should the sign of AV be for r < R? (i.e. if we start at r = R and move in, will V go down or up, or does it do one and then the other?) Should V increase or decrease with r (If we start at r = 0 and move out, what does V do?) Check that your V graph is consistent with these. (c) Taking the electric potential at the sphere's surface to be equal to zero (ie setting our starting point at the surface, as we did in the graphs,) integrate the field as a function of r to find an expression (in terms of Q, R, and r) for the change in electric potential AV(r) between the surface and test points at a radial distance r ≤ R from the sphere's center (i.e. inside.) Check that AV = 0 when r = R! Hint: you may need to think twice about what your integration variable and limits are. A diagram helps. Make sure you use the correct function for E(r) in this region. (d) Again, taking the electric potential at the sphere's surface to be equal to zero, find an expression (in terms of Q, R, and r) for the electric potential difference AV(r) between the surface and points at a radial distance r > R from the sphere's center (i.e. outside.) Check that AV = 0 when r = R! Make sure you use the correct function for E(r) in this region. Check that your results agree with your plot of V(r). (e) How would your function for V(r) and plot change if the same excess charge Qwere placed on a conducting sphere of the same radius R? How would your function for V(r) and plot change if we set V = 0 at infinity?
Transcribed Image Text:by using positive values for field pointing out and negative for field pointing in, but since the field all points out everywhere, we don't have to worry about that. Taking the electric potential at the sphere's surface (at r = R) to be equal to zero (ie setting our starting point at the surface,) draw a qualitatively correct graph of V(r), the potential as a function of radial distance from the center. Note that we DO need to be careful about the sign of V! The idea in this part is to draw your graph using the information in the E graph, but NOT do any calculations. Check that your V graph and E graph are consistent with each other at r = 0, at r < R, at r = R, and at r > R. Also check that your V graph is consistent with our choice of V=0 at r=R. (b) What should the sign of AV be for r > R? Should V increase or decrease with r (i.e. if we start at r = R and move out, will V go down or up, or does it do one and then the other? How did you decide? What should the sign of AV be for r < R? (i.e. if we start at r = R and move in, will V go down or up, or does it do one and then the other?) Should V increase or decrease with r (If we start at r = 0 and move out, what does V do?) Check that your V graph is consistent with these. (c) Taking the electric potential at the sphere's surface to be equal to zero (ie setting our starting point at the surface, as we did in the graphs,) integrate the field as a function of r to find an expression (in terms of Q, R, and r) for the change in electric potential AV(r) between the surface and test points at a radial distance r ≤ R from the sphere's center (i.e. inside.) Check that AV = 0 when r = R! Hint: you may need to think twice about what your integration variable and limits are. A diagram helps. Make sure you use the correct function for E(r) in this region. (d) Again, taking the electric potential at the sphere's surface to be equal to zero, find an expression (in terms of Q, R, and r) for the electric potential difference AV(r) between the surface and points at a radial distance r > R from the sphere's center (i.e. outside.) Check that AV = 0 when r = R! Make sure you use the correct function for E(r) in this region. Check that your results agree with your plot of V(r). (e) How would your function for V(r) and plot change if the same excess charge Qwere placed on a conducting sphere of the same radius R? How would your function for V(r) and plot change if we set V = 0 at infinity?
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