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PHYS 2420 Electric Fields from Charge Distributions Solutions

1. Find the electric field at a point a distance a from one end of a long thin wire of length L
   (a) if it has total charge Q. Examine the limit a >> L and show that your result is identical to that of a point charge.
   (b) if the charge distribution was λ(x) = 2Qx/L².

   

   To calculate the electric field of an object at a point outside the object (the field point) from first principles we need to do the following:

   • consider the object to be made up of small pieces each carrying charge dq,

   • set up and origin and coordinate system,

   • label the vector distance S from the origin to the small piece of charge (the source point) and the vector distance P from the origin to where the field is to be calculated (the field point),

   • label the vector distance, r, from the source point to the field point. Note r = P – S,

   • express r in terms of the coordinate system and P,

   • express the charge in terms of the coordinate system using dq = ρdV or dq = ΣdA or dq = λdL,

   • use Coulomb’s Law, i.e. dE = kdqr/r³,

   • integrate over the entire volume or area or length as appropriate.

   (a) The diagram below shows the origin and coordinate system that I have chosen.

   

   The electric field points to the right and is

   dE = rkdq/r³ .

   The vector distances reduce to S = ix, P = i(L + a), and r = i r. The distance from source to point is r = L + a – x. Therefore the electric field is

   dE = i kλdx/(L + a – x)²

   We need to integrate over the entire length of the object from x = 0 to x = L. The integral for the total field is therefore

   E = i ò₀^L kλdx/(L + a – x)² .

   Evaluating the integral yields

   E = i KQ / a(L+a).

   In the limit a >> L, a(L + a) ® a². Thus the electric field strength is E = KQ/a² as required.

   (b)The only modification from part a) is that the charge density is a function of x, so our integral is

   

   In the limit a >> L, aln(a / L + a) ® –L + ½(L/a). Thus the electric field strength is E = KQ/a² as required.

   

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2. Find the electric field at a point a distance a from the centre of a long thin wire of length L and total charge Q. Show that your result reduces to that of a point charge in the limit a >> L. Also show that your answer reduces to E = 2kλ/a in the limit L >> a, the well–known and very useful result for a long thin wire. Note λ = Q/L.

   

   The diagram below shows the origin and coordinate system that I have chosen.

   

   The electric field is

   dE = r kdq/r³ .

   Note that r and dE have both x and y components. The vector distance S reduces to S = ix while P = ja. Thus r = P – S = –i x + j a. The distance from source to field point is given by r = (a² + x²)^½. Therefore the components of the electric field are

   dE = –i dE_x + j dE_y = –i kλxdx/(a² + x²)^3/2 + j kλadx/(a² + x²)^3/2 .

   We need to integrate over the entire length of the object from x = –½L to x = ½L. Paying attention to the symmetry of the integrals, we see that the x component must vanish. The net electric field can only point upwards. The integral for the total field is therefore

   E = j ò_–½L^½L kλadx/(a² + x²)^3/2 .

   Evaluating the integral yields

   E = j Kl L /a(a²+¼L²)^½.

   In the limit a >> L, (a²+¼L²)^½ ® a. Thus the electric field strength becomes E = K(l L)/a², or E = KQ/a² since Q = l L. When one is very far from the rod its field appears to be that of a point source.

   In the limit L >> a, (a²+¼L²)^½ ® L. Thus the electric field strength approaches E = 2Kl /a as required.

   

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3. A wire has been bent into a semicircle of radius R. It has a linear charge density λ. Determine the electric field at point P, at the centre of the circle.

   

   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.

   

   The electric field is

   dE = r kλdx/r³ = r kλRdθ/r³

   Note that r and dE have both x and y components. The source point is given by S = i cos(θ) + j sin(θ). The field point is P = 0. Thus r = P – S = –i cos(θ) – j sin(θ). The distance from source to field point is given r = R. Therefore the components of the electric field are

   dE = –i dE_x + j dE_y = –i (kλ/R) cos(θ) dθ – j (kλ/R) sin(θ) dθ .

   The integral is evaluated from θ = 0 to θ = π. Paying attention to the symmetry of the integrals, we see that the x component must vanish. The net electric field can only point downwards. The integral for the total field is therefore

   E = –j ò₀^p (kλ/R) sin(θ) dθ .

   Evaluating the integral yields

   E = j 2kl /R .

   

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4. Find the electric field at a point a distance h from the centre of a circular plate of radius R with charge Q. Show that your answer reduces to that of a point charge in the limit h >> R (the point is very far away from the plate). Show that your answer reduces to the well–known result E = Σ/2ε₀ when R >> h (the plate is infinite or the point is very close to the plate). Note Σ = Q/A.

   

   We will assume that the plate is on the x–y plane and that h is in the positive z direction. The plate is circular which suggests the use of polar coordinates for the variable of integration. The diagram below shows the origin and coordinate system that I have chosen.

   

   The electric field is

   dE = a kdq/a³ = a kΣrdrdθ/a³ ,

   where s = Q/A = Q/πR². Note that a and dE have both x, y, and z components. The source point is given by S = i rcos(θ) + j rsin(θ) + k 0. The field point is P = i 0 + j 0 + z h. Thus a = P – S = –i rcos(θ) – j rsin(θ) + z h. The distance from source to field point is given a = (r² + h²)^½. Therefore the components of the electric field are

   dE	= –i dEx + j dEy + k dEx
   	= –i (kΣr2drcos(θ)dθ)/(r2+h2)3/2 – j (kΣr2drsin(θ)dθ)/(r2+h2)3/2 + k (kΣhrdrdθ)/(r2+h2)3/2 .

   We need to integrate over the entire area of the object from r = 0 to r = R and θ = 0 to θ = 2π. Paying attention to the symmetry of the integrals, we see that the x and y components must vanish. The net electric field can only point in the z direction. The integral for the total field is therefore

   E = k ò₀^R ò₀^2p (kΣhr dr dθ)/(r²+h²)^3/2 .

   Evaluating the integral yields

   E = k 2πkΣ{1 – h/(h² + R²)^½} .

   In the limit of h >> R, 1 – h/(h² + R²)^½ ® ½R²/h². Thus the strength of the field becomes E = k πR²s /h² = kQ/h², the same as a point charge.

   In the limit where the field point is very close to the plate R >> h, h/(h² + R²)^½ ® 0 and the field becomes E = 2πkΣ. Recalling that k = 1/4pe₀, we thus have E = Σ/2ε₀ as required.

   

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

   

   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 ) .

   he magnitude of r is

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

   The small portion of electric field is

   dE = kdq r / r³

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

   

   The i and j components will vanish when we do the integral over f . This is in accord with symmetry considerations. The k component of the integral over f gives 2p since the integrand has no f dependence. We are thus left with

   

   For the case when a > R, this reduces to E_z = (s /2e ₀)(R/a)². Actually, by symmetry, this applies in any direction.

   For the case when a < R, this reduces to E_z = 0. Actually, by symmetry, this applies in any direction inside the sphere.

   

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6. 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 field 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 clearcut 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 electric field is

   dE = kdq r / r³

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

   

   Now the i and j contributions must cancel by symmetry. This can be confirmed by noting that the integrand for each is odd and we are integrating over an even interval. Thus field is 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 field is

   dE = kdq r / r³

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

   

   The i components will vanish when we do the integral over q while the k components will vanish when we do the integral over z. This is in accord with symmetry considerations. We are thus left with

   

   

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7. The solid half–cylinder shown above has charge density r . It has height L and radius R. Derive the integral necessary to find the electric field 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 = r 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 dE = kdq G / G³ which we need to integrate over the entire surface of the half-cylinder.



The i components will vanish when we do the integral over q while the k components will vanish when we do the integral over z. This is in accord with symmetry considerations. We are thus left with

Questions? mike.coombes@kwantlen.ca