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PHYS 2420 Electric Potential from Charge Distributions Solutions

1. A rod is bent into a semi–circular arc of radius R. The rod has a uniform linear charge distribution λ. Find the potential at a distance y below the centre of the arc. Find the component of the electric field in the y direction.

   We calculate the potential by direct integration in much the same way that we calculate the electric field. The exception is that potential is a scalar quantity and that dV = k dq/r.

   The shape of the object suggests the use of polar coordinates for the integration variables. The diagram below shows the origin and coordinate system that I have chosen.

   

   Note that r has both x and y components. The source point is given by S = i Rcos(θ) + j Rsin(θ). The field point is P = –j y. Thus r = P – S = – i Rcos(θ) – j [Rsin(θ) + y]. The distance from source to field point is given r = |r| = {R² + y² + 2Rysin(q )}^½. Therefore the potential

   dv = kl Rdq /{R² + y² + 2Rysin(q )}^½ .

   The integral is evaluated from θ = 0 to θ = π. The integral for the total potential is therefore

   V = kλR ò₀^p dq / {R² + y² + 2Rysin(q )}^½ .          (1)

   Unfortunately the integral does not have a simple solution but may be solved numerically if desired.

   The gradient along the y direction is E_y = –j dV/dy. Taking the derivative of equation (1) yields

   E_y = j kλR ò₀^p dθ (y + Rsin(q ))/{R² + y² + 2Rysin(q )}^3/2 .

   If we take the limit as y goes to zero, we find

   E_y = j (kl/R) ò₀^p dθ sin(q ) = j 2kl/R .

   As we had before.

   

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2. Find the potential at the side of a circular plate a distance a from the its centre. The plate has a uniform charge distribution s . Determine the electric field along a.

   

   We calculate the potential by direct integration in much the same way that we calculate the electric field. The exception is that potential is a scalar quantity and that dV = kdq/r.

   The shape of the object suggests the use of polar coordinates for the integration variables. The diagram below shows the origin and coordinate system that I have chosen.

   

   Note that b has both x and y components. The source point is given by S = i rcos(θ) + j rsin(θ). The field point is P = i a. Thus b = P – S = i [a – rcos(θ)] + j rsin(θ). The distance from source to field point is given b = |b| = {r² + a² – 2arcos(q )}^½. Therefore the potential

   dV = ks rdrdq /{r² + a² – 2arcos(q )}^½ .

   The variables of integration range from r = 0 to r = R and q = 0 to q = 2π . Therefore the total potential is

   V = ks ò₀^2π dq ò₀^R dr r /{r² + a² – 2arcos(q )}^½.

   Again the integral is messy to try and solve.

   

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3. A hollow sphere of radius R has a surface charge s . Find the electric potential at a point along the z axis outside the sphere. Find the electric potential at a point along the z axis inside the sphere. Find the electric field.

   

   In both cases the setup is the same. We select a tiny piece of the surface which has charge dq = s R²sinq dq df . We assume it is at some position S = ix + jy + kz. Expressing x, y, and z in spherical coordinates, we have

   S = iR sinq cosf + jR sinq sinf + kR cosq .

   Hence, r = P – S becomes

   r = –iR sinq cosf – jR sinq sinf + k(a – R cosq ) .

   The magnitude of r is

   |r|² = |P|² + |S|² – 2P•S = a² + R² – 2aR cosq .

   The small portion of electric potential is

   dV = kdq / r

   which we need to integrate over the entire surface of the sphere.

   

   The integral over f gives 2π . The integral over q is exact but we must be careful of the two cases. We find

   

   The potential depends only on the distance from the centre of the shell. To find the electric field, we use E = –ÑV which reduces to E = a ∂V/∂a in the case of radial symmetry. Here a is the unit radial vector. We find (after changing a to r)

   

   

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4. A hollow half cylinder is shown below has surface charge s . The cylinder has height L and radius R. The back of the hollow cylinder is a flat rectangle. Derive the integrals necessary to find the electric potential outside the cylinder on the y axis a distance d from the back of the half cylinder. There is no need to solve the integrals.

   

   There is no clear-cut symmetry here. The key to solving this problem is to break the object into two simpler pieces, with obvious symmetry, as shown below

   

   The flat back piece exhibits Cartesian symmetry and the front piece has cylindrical symmetry. We do the back first.

   

   We select a tiny piece of the surface which has charge dq = s dxdz. We assume it is at some position S = ix + kz. The field point is P = jd.

   Hence, r = P – S becomes

   r = –ix + jd – kz .

   The magnitude of r is

   r = |r| = [x² + d² + z²]^½ .

   The small portion of the electric potential is

   dV = kdq/r

   which we need to integrate over the entire surface of the plate. Thus

   

   Next we consider the cylindrical piece.

   

   We select a tiny piece of the surface which has charge dq = s Rdq dz. We assume it is at some position S = ix + jy + kz. Expressing x, y, and z in cylindrical coordinates, we have

   S = i R cosq + j R sinq + k z .

   Hence, r = P – S becomes

   r = –i R cosq + j (d – R sinq ) – k z .

   The magnitude of r is

   r = |r| = d² + R² + z² – 2dR sinq .

   The small portion of electric potential is

   dV = kdq/r

   which we need to integrate over the entire surface of the half-cylinder.

   

   

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5. The solid half–cylinder shown above has charge density ρ . It has height L and radius R. Derive the integral necessary to find the electric potential at a point outside the cylinder on the y axis. There is no need to solve the integral.



We select a tiny piece of the volume which has charge dq = ρ rsinq drdq dz. We assume it is at some position S = ix + jy + kz. Expressing x, y, and z in cylindrical coordinates, we have

S = i r cosq + j r sinq + k z .

Hence, G = P – S becomes

G = –i r cosq + j (d – r sinq ) – k z .

The magnitude of G is

G = |G| = d² + r² + z² – 2dr sinq .

The small portion of electric field is dV = kdq/G which we need to integrate over the entire surface of the half-cylinder.





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6. The following diagram shows the electric field due to a radially symmetric charge distribution. Deduce the charge distribution.

   

   The fact that the electric field drops as 1/r² for large r indicates that we are dealing with spherical symmetry not cylindrical. Notice the jumps in E at r = R, r = 2R, and r = 3R. This only occurs when we cross a shell of charge at each radius. There is no charge between the shells since again the field drops as 1/r². There is obviously no charge inside the first shell since E is zero. The first shell must have charge Q. Now Gauss' Law for spherical charge distributions reduces to

   E(r) = Q_inside / 4πε₀r² .          (1)

   At r = 2R, Q_inside = Q₁ + Q₂ = Q + Q₂. Also from the graph we know, E(2R) = Q/4πε₀R². Substituting this information in Eq. (1), we have

   Q/4πε₀R² = (Q + Q₂)/4πε₀(2R)² ,

   or, eliminating common factors,

   Q = (Q + Q₂)/4 .

   We thus have Q₂ = 3Q.

   We repeat this approach at the next shell. At r = 3R, Q_inside = Q₁ + Q₂ + Q₃ = 4Q + Q₃. Also from the graph we know, E(3R) = Q/4πε₀R². Substituting this information in Eq. (1), we have

   Q/4πε₀R² = (4Q + Q₃)/4πε₀(3R)² ,

   or, eliminating common factors,

   Q = (4Q + Q₃)/9 .

   We thus have Q₃ = 5Q.

   

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7. The following diagram shows the electric field due to a cylindrical charge distribution. Deduce the charge distribution.

   

   The fact that there is a drop in the electric field at r = 2R indicates a charged cylindrical shell at that point. The fact that the electric field drop to zero indicates that the total charge between 0 and 2R is zero from Gauss' Law. The fact that the field drops as 1/r between R and 2R indicates a vacuum, as this is the typical drop in the electric field as you move away from a long charged cylinder. The fact that the field rises between 0 and R tells us that the charge is spread out over the volume of the cylinder. Gauss' Law for uniform charge distributions is

   E(r) = Q_inside/2πε₀rL ,

   usually written

   E(r) = l/2πε₀r ,

   where l = Q_inside/L. Now

   Q_inside = ò _inside ρ dV = 2πL ò ₀^r ρ(a) a da ,

   where a is our radial variable of integration. Thus we have

   E(r) = (1/ε ₀r) ò ₀^r ρ (a) a da .          (2)

   We are trying to find a form of ρ (a) that will yield are r² behaviour in E. The presence of 1/r outside the integral, indicates that the integral must produce an r³ result. This in turn requires that ρ (a) = Ca where C is a constant. Using this form we evaluate Eq. (2) and find

   E(r) = Cr²/3ε₀ .

   We know E(R) = l/2πε₀R, therefore C = 3l/2πR³ and ρ(r) = 3lr/2πR³. The outside shell must have linear charge density –l.

   

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8. The following diagram shows the electric potential due to a radially symmetric charge distribution. If there is a volume charge in any region, the charge density ρ is constant. Deduce the charge distribution.

   

   The fact that the potential drops as 1/r for large r indicates that we are dealing with spherical symmetry not cylindrical. Notice the change in the slope of V at r = 2R. This only occurs when we cross a shell of charge at that radius. There is also a change in the slope of V at r = R so there is surface charge there as well. The flat region from R to 2R of constant potential occurs for volumes with zero electric field so this regions is uncharged space. The rising potential from r = 0 to r = R indicates some charge distribution. Thus what we have is a sphere of radius R with unknown charge density ρ . On the surface of that sphere is some unknown charge density s_R. Then there is a gap followed by a shell with surface density s_2R as shown below.

   

   We know that the potential outside any charged sphere with radial symmetry is V = Q_total/4πε₀r. At r = 2R, the graph says V = Q/8πε₀R. So here the total charge enclosed must be Q. Therefore, using Gauss' Law

   Q_sphere + s_R4πR² + s_2R4π(2R)² = Q .           (1)

   Inside R, the potential must be given by

   V = Cr² .          (2)

   To find the constant C, we compare Eq. (2) with the value of V at r = R from the graph

   CR² = Q/8πε₀R .          (3)

   We find that C = Q/8πε₀R³. Hence

   V(r) = Qr²/8πε₀R³ .          (4)

   To find Q_sphere, s_R, and s_2R, we need to use the boundary conditions at R and 2R. At R,

   ∂V/∂ r|_R+ – ∂V/∂r|_R- = s_R/ε₀ .

   Above r = R, we know ∂V/∂r|_R+ = 0 since the potential is flat. Below r = R, ∂V/∂r|_R- = Q/4πε₀R² using Eq. (4). Hence s_R = -Q/4πR². Repeating for r = 2R,

   ∂V/∂r|_2R+ – ∂V/∂r|_2R- = s_2R/ε₀ .

   Above r = 2R we know from the graph that V(r) = Q/4πε_or. Thus ∂V/∂r|_2R+ = –Q/16πε₀R². Below r = 2R, the potential is flat and ∂V/∂r|_2R- = 0. Thus s_2R = Q/16πR². We can now find Q_sphere by substituting these results back into Eq. (1),

   Q_sphere + (-Q/4πR²)4πR² + (Q/16πR²)4π(2R)² = Q .           (5)

   Hence Q_sphere = Q. The inner shell has charge - Q and the outer shell has charge +Q. The charge density in the sphere is thus ρ = Q/(4πR³/3).

   

Questions? mike.coombes@kwantlen.ca