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PHYS 2420 Electric Potential from Electric Fields Solutions

1. The electric field due to a charged spherical shell is given by

   

   Determine V as a function of a where a is the distance from the centre of the sphere. Plot V as a function of a.

   The electrostatic potential when the electric field is known is given by

   V = –ò_O^B E·dL .

   When we are asked for the potential of a point, we must remember that potential is always relative to some arbitrary reference position here labelled O. The reference point should be a point where the electric field is zero. The most common reference point is at infinity. We will use that reference here. Therefore we define the potential as

   V = –ò_¥^r=a E·dL .

   We are given the form of the electric field. In the diagram below the electric field and one possible path from infinity to a particular point at r = a is shown.

   

   The electric field is radial, that is E(r) = E(r) r. The path element is dL = rdr + qrdθ + fsin(θ)dφ where r, q, and f are the spherical coordinate unit vectors. Therefore the potential reduces to

   V = - ò_¥^a E(r) dr .

   The electric field looks like

   

   Since there is a discontinuity if the electric field at r = R, we need to consider two cases.

   (i) The first is when a > R. Here we know the form of E(r) and we find

   V = –ò_¥^a kQ/r² dr = kQ/r |_¥^a = kQ/a .

   (ii) In the second case 0 < a < R and we have to break up the integral for the two regions.

   V = –ò_R^a E(r) dr – ò_¥^R E(r) dr = – ò_¥^R E(r) dr .

   Note that the first integral is zero since E = 0 inside the sphere. We are given the form of E(r) outside the sphere, so we find

   V = – ò_¥^R kQ/r² dr = kQ/r |_¥^R = kQ/R .

   So the potential is constant everywhere inside the sphere.

   The potential looks like

   

   Note that there is only a change in slope for V at the boundary unlike E.

   

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2. The electric field due to a large thin plate is given by

   

   Determine V as a function of a where a is the distance from the centre of the plate. Plot V as a function of a.

   The electrostatic potential when the electric field is known is given by

   V = –ò_O^B E·dL .

   When we are asked for the potential of a point, we must remember that potential is always relative to some arbitrary reference position here labelled O. The reference point should be a point where the electric field is zero. The most common reference point is at infinity but in this case E is not zero at infinity. We will use x = 0 instead. Therefore we define the potential as

   V = –ò₀^x=a E·dL .

   We are given the form of the electric field and the electric field has only an x component, that is E(r) = iE(r). The path element is dL = idx + jdy + kdz. Therefore the potential reduces to

   V = - ò₀^a E(x) dx .

   The electric field looks like

   

   Since there is a discontinuity if the electric field at x = - ½d and x = ½d, we need to consider three cases.

   (i) The first is when - ½d £ a £ ½d. Here we know the form of E(x) and we find

   V = - ò₀^a E(x) dx = - ò₀^a (ρ₀/3e₀) x³ dx = - (ρ₀/12e₀) x⁴ |₀^a = - ρ₀a⁴/12e ₀ .

   (ii) In the second case a > ½d and we have to break up the integral for the two regions.

   V	= - ò0 ½d E(x) dx - ò½da E(x) dx
   	= - ò0 ½d (ρ0/3e 0) x3 dx - ò½da(ρ0d3/24e 0) dx
   	= - (ρ0/12e 0) x4 |0½d - (ρ0d3/24e 0) x |½da
   	= - (ρ0d4/192e 0) - (ρ0d3a /24e0) + (ρ 0d4 /48e0)
   	= (ρ0d4/64e0) - (ρ0d3a /24e0)

   Note that the potential has the same value on either side of the discontinuity at a = ½d.

   (iii) In the third case a < - ½d and we have to break up the integral for the two regions.

   V	= - ò0 - ½d E(x) dx - ò-½da E(x) dx
   	= - ò0 - ½d (ρ0/3e0) x3 dx - ò-½da -(ρ0d3/24e0) dx
   	= - (ρ0d3/24e0) x4 |0½d + (ρ0d3/24e0) x |-½da
   	= - (ρ0d4/192e0) + (ρ 0d3a /24e0) + (ρ0d4 /48e0)
   	= (ρ0d4/64e0) + (ρ0d3a /24e0)

   Note that the potential has the same value on either side of the discontinuity at a = - ½d.

   The potential looks like

   

   

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3. The electric field for a round flat plate of charge density s is

   

   Determine V as a function of a where a is the distance from the centre of the plate. Plot V as a function of a.

   The electrostatic potential of a point when the electric field is known is given by

   V = –ò_O^B E·dL .

   When we are asked for the potential of a point, we must remember that potential is always relative to some arbitrary reference position here labelled O. The reference point should be a point where the electric field is zero. The most common reference point is at infinity and that will be our choice here. Therefore we define the potential as

   V = –ò_¥^a E·dL .

   We are given the form of the electric field and the electric field has only an x component, that is E(r) = iE(r). The path element is dL = idx + jdy + kdz. Therefore the potential reduces to

   V = - ò_¥^a E(x) dx .

   The electric field looks like

   

   Since there is a discontinuity if the electric field at x = 0, we need to consider two cases.

   (i) The first is when a > 0. Here we know the form of E(x) and we find

   V	= –ò¥a (s/e0) {1 – x/(x2 + R2)½}dx
   	= –(Σ/2ε0){x – (x2 + R2)½}| ¥a
   	= (Σ/2ε0){(a2 + R2)½ – a}

   (ii) In the second case a < 0. Again we know the form of E(x) and we find

   V	= –ò¥a - (s/e0) {1 + x/(x2 + R2)½}dx
   	= (Σ/2ε0){x + (x2 + R2)½}| ¥a
   	= (Σ/2ε0){(a2 + R2)½ + a}

   The potential looks like

   

   

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4. The electric field due to a large plate was given in the previous question. A capacitor consists of two identical plates with equal but opposite charge distributions.
   (a) Find the net field just to the right of the centre of the capacitor. Do the same on the left side. Find the net electric field at a point between the place a distance a from the positive plate. Show that is R >> d that the E_net = 0 outside the plates and E_net = Σ/e₀ between the plates.
   (b) Find the potential difference between the plates. Show that the potential difference between the plates is directly proportional to the separation d of the plates when R >> d.

   (a) The diagram below shows the electric field of each plate each in a different colour. A coordinate system has been set up with the origin at the left–hand positive plate.

   

   We know that E(x) = (Σ/2ε₀){1 – h/(h² + R²)^½} where h is the distance from the plate.

   (i) Just outside the left plate.

   The electric field due to the positive plate will be E₊ = –i Σ/2ε₀ since h = 0. The electric field due to the negative plate will be E_– = i (Σ/2ε₀){1 – d/(d² + R²)^½}. Hence the net field will be

   E_right = E₊ + E_– = –i (Σ/2ε₀) d/(d² + R²)^½ .

   (ii) Just outside the right plate.

   The electric field due to the positive plate will be E₊ = i Σ/2ε₀{1 – d/(d² + R²)^½} since h = d. The electric field due to the negative plate will be E_– = –i (Σ/2ε₀) since h = 0. Hence the net field will be

   E_left = E₊ + E_– = i (Σ/2ε₀) d/(d² + R²)^½ .

   (iii) At point a between the plates.

   The electric field due to the positive plate will be E₊ = i Σ/2ε₀{1 – a/(a² + R²)^½} since h = a. The electric field due to the negative plate will be E_– = i (Σ/2ε₀) {1 – [d–a]a/([d–a]² + R²)^½} since h = d–a. Hence the net field will be

   E_between = E₊ + E_– = –i (Σ/ε₀){1 – ½ a/(a² + R²)^½ – ½[d–a]/([d–a]² + R²)^½} .

   The exterior fields are proportional to d/R and thus will vanish for large plates. Inside the capacitor a/R and [d–a]/R are both small and thus E_between = Σ/ε₀ for large plates.

   (b) The potential difference between the plates is given by

   V = ò_d⁰ E_between dx ,

   where we need to substitute x for a in our formula for E_between. Thus our integral is

   V = ò_d⁰ (Σ/ε₀){1 – ½ x/(x² + R²)^½ – ½[d–x]/([d–x]² + R²)^½} dx .

   Doing the integral using MAPLE, we find

   V = Σd/ε₀ – (Σ/ε₀){ (d² + R²)^½ – R }.

   The second term is proportional to (d/R)² and thus will vanish for large R. So for large plates the potential is directly proportional to d, V = Σd/ε₀.

   

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5. The electric field due to a very long cylinder is given by

   

   Determine V as a function of a where a is the distance from the centre of the cylinder. Plot V as a function of a.

   The electrostatic potential when the electric field is known is given by

   V = - ò_O^B E·dL .

   When we are asked for the potential of a point, we must remember that potential is always relative to some arbitrary reference position here labelled O. The reference point should be a point where the electric field is zero. The most common reference point is at infinity. In cylindrical case, this is usually a bad choice since the field does not drop off fast enough. We will O = 0 instead. Therefore we define the potential as

   V = - ò₀^r=a E·dL .

   We are given the form of the electric field. The electric field is radial, that is E(r) = E(r) r. The path element is dL = rdr + qrdθ + zsin(θ)dz where r, q, and z are the cylindrical coordinate unit vectors. Therefore the potential reduces to

   V = - ò₀^a E(r) dr .

   The electric field looks like

   

   Since there is a discontinuity if the electric field at r = R, we need to consider two cases.

   (i) The first is when 0 £ a £ R. Here we know the form of E(r) and we find

   V = - ò₀^a (Br³/4e₀) dr = - (Br⁴/16e₀) |₀^a = - (Ba⁴/16e₀).

   (ii) In the second case a > R and we have to break up the integral for the two regions.

   V	= - ò0 R E(r) dr - òRaE(r) dr
   	= - ò0 R (Br3/4e0) dr - òRa (BR4/4e0r)dr
   	= - (Br4/16e0) |0a - (BR4/4e0) ln(r) |Ra
   	= - (BR4/16e0) - (BR4/4e0) ln(a/R)

   Note that the potential has the same value on either side of the discontinuity at a = R.

   The potential looks like

   

   

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6. The electric field due to a spherical charge distribution is given by

   

   Determine V as a function of a where a is the distance from the centre of the spherical charge distribution. Plot V as a function of a.

   The electrostatic potential when the electric field is known is given by

   V = - ò_O^B E·dL .

   When we are asked for the potential of a point, we must remember that potential is always relative to some arbitrary reference position here labelled O. The reference point should be a point where the electric field is zero. The most common reference point is at infinity. We will use that here. Therefore we define the potential as

   V = - ò_¥^a E·dL .

   We are given the form of the electric field. The electric field is radial, that is E(r) = E(r) r. The path element is dL = rdr + qrdθ + fsin(θ)dφ where r, q, and f are the spherical coordinate unit vectors. Therefore the potential reduces to

   V = - ò_¥^a E(r) dr .

   The electric field looks like

   

   Since there is a discontinuity if the electric field at r = R and r = 2R, we need to consider three cases.

   (i) The first is when a ³ 2R. Here we know the form of E(r) and we find

   V = - ò_¥^a (Q/4π e₀r²) dr = (Q/4π e₀r) |_¥^a = Q/4π e₀a.

   (ii) In the second case R £ a £ 2R and we have to break up the integral for the two regions.

   V	= - ò¥ 2R E(r) dr - ò2RaE(r) dr
   	= - ò¥ 2R (Q/4π e0r2) dr - ò2Ra (Q/28π e0)(r/R3 – 1/r2)dr
   	= (Q/4π e0r) |¥ 2R - (Q/28π e0)(r2/2R3 + 1/r)|2Ra
   	= (Q/8π e0R) - (Q/28π e0)(a2/2R3 + 1/a) + (5Q/56π e0R)
   	= (3Q/14π e0R) - (Q/28π e0)(a2/2R3 + 1/a)
   	= (Q/56π e0R)[12 – (a/R)2 – 2(R/a)]

   Note that the potential has the same value on either side of the discontinuity at a = 2R.

   (iii) In the third case 0 £ a £ R and we have to break up the integral for the three regions.

   V	= - ò¥ 2R E(r) dr - ò2RRE(r) dr - òRaE(r) dr
   	= - ò¥ 2R (Q/4π e0r2) dr - ò2RR (Q/28π e0)(r/R3 – 1/r2)dr – 0
   	= (Q/4π e0r) |¥ 2R - (Q/28π e0)(r2/2R3 + 1/r)|2RR
   	= (Q/8π e0R) - (3Q/56π e0R) + (5Q/56π e0R)
   	= 9Q/56π e0R

   Note that the potential has the same value on either side of the discontinuity at a = R.

   The potential looks like

   

   

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7. The electric field due to a spherical charge distribution is given by

   

   Determine V as a function of a where a is the distance from the centre of the spherical charge distribution. Plot V as a function of a.

   The electrostatic potential when the electric field is known is given by

   V = –ò_O^B E·dL .

   When we are asked for the potential of a point, we must remember that potential is always relative to some arbitrary reference position here labelled O. The reference point should be a point where the electric field is zero. The most common reference point is at infinity. We will use that reference here. Therefore we define the potential as

   V = –ò_¥^a E·dL .

   We are given the form of the electric field. The electric field is radial, that is E(r) = E(r) r. The path element is dL = rdr + qrdθ + fsin(θ)dφ where r, q, and f are the spherical coordinate unit vectors. Therefore the potential reduces to

   V = - ò_¥^a E(r) dr .

   The electric field looks like

   

   Since there is a discontinuity if the electric field at r = R, we need to consider two cases.

   (i) The first is when a > R. Here we know the form of E(r) and we find

   V = –ò_¥^a (AR⁵/5e₀r²) dr = (AR⁵/5e₀r) |_¥^a = AR⁵/5e₀a .

   (ii) In the second case 0 < a < R and we have to break up the integral for the two regions.

   V	= - ò¥ R E(r) dr - òRaE(r) dr
   	= - ò¥ R (AR5/5e0r2) dr - òRa (Ar3/5e0)dr
   	= (AR5/5e0r) |¥ R - (Ar4/20e0)|Ra
   	= (AR4/5e0) - (A/20e0) (a4 – R4)
   	= (A/4e0)(R4 – a4/5)

   Note that the potential has the same value on either side of the discontinuity at a = 2R.

   The potential looks like

   

   

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8. The electric potential over a certain region is given by V = 3x²y–4xz–5y² volts. Determine the Cartesian components of the electric field and evaluate at the point (+1,0,+2).

   The electric field is related to the potential by E = –ÑV. In Cartesian coordinates the relationship is

   E = –i ∂V/∂x – j ∂V/∂y – k ∂V/∂z .

   For this particular case we find

   E = –i (6xy – 4z) – j (3x² – 10y) – k (–4x) .

   Evaluating at the point of interest, we find E = 8i – 3j + 4k.

   

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9. Over a certain region of space, the electric potential is V = 5r³cos(q)sin²(f). Determine the spherical components of the electric field and evaluate electric field at the point (r = 1, q = π/2, f = π /4).

   The electric field is related to the potential by E = –ÑV. In spherical coordinates the relationship is

   E = –r ∂V/∂r – q (1/r)∂V/∂q – f (1/rsin(q))∂V/∂f .

   For this particular case we find

   E = –r [15r²cos(q)sin²(f)] + q [5r²sin(q)sin²(f)] – f [10r²sin(f)cos(f)/tan(q)].

   Evaluating at the point of interest yields E = –r 0 – q 5/2 – f 0.

   

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10. The potential in a certain region of space is given by V = Ar. Determine the charge density ρ . Determine the total charge in a sphere of radius R centred at r = 0.

    The electrostatic potential and the charge density are related through Poisson’s Equation, ѲV = –ρ /e₀. In spherical coordinates, Poisson’s Equation is defined

    

    Doing the required derivatives, we find ρ = –2e₀A/r.

    The total charge is defined as

    Q = ò _sphere ρ dV = ò ₀^R ò ₀^π ò₀^2π ρ r²dr sin(q) dq df = 4π ò₀ ^R [–2e₀Ar] dr = –4π e₀AR² .

    

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11. In a semiconducting medium, the potential of a charge q is found to be

    

    Find the corresponding electric field and charge density.

    The electric field is related to the potential by E = –ÑV. In spherical coordinates the relationship is

    E = –r ∂V/∂r – q (1/r)∂V/∂q – f (1/rsin(q))∂V/∂f .

    For this particular case we find

    E = r (q e^-r/l /4π e₀)(1/r² + 1/l r) .

    The electrostatic potential and the charge density are related through Poisson’s Equation, ѲV = –ρ/e₀. In spherical coordinates, Poisson’s Equation is defined

    

    Doing the required derivatives, we find

    

    

Questions? mike.coombes@kwantlen.ca