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PHYS 2420 Electric Energy & Method of Images Solutions

1. Charges are arranged as shown below. What is the electrostatic potential energy of each configuration?
   (a) The side of the square has length L.

   

   The separation of the charges is L.

   

   Electrostatic energy of a charge Q₁ with respect to another charge is Q₀ is KQ₁Q₀/r where is the distance between the two charge. This is the energy it takes to do work against the electric field of Q₀ as we bring Q₁ in from infinity. We can imagine that the charges are brought to the current position one at a time from very far away. Bringing in the first charge requires no work as there is no external field to act on our charge. Bringing in the second charge means we have to act against the field of the first charge. Bringing in the third charge means we have to work against the fields of both the first and second charges, and so on. Formulated mathematically the energy is

   .

   (a) Recalling that the distance across a diagonal is Ö2L and taking the upper left charge to be charge #1 and moving clockwise, we find

   U	=	K(Q)(–Q)/L + K(Q)(Q)/(Ö2L) + K(Q)(–Q)/L + K(–Q)(Q)/L + K(–Q)(–Q)/(Ö2L) + K(–Q)(Q)/L
   	=	–(4 – Ö2) KQ2/L

   (b) Taking the leftmost charge as #1, we find

   U	=	K(–Q)(Q)/L + K(Q)(Q)/(2L) + K(Q)(–Q)/L + K(–Q)(Q)/(3L) + K(–Q)(–Q)/(2L) + K(–Q)(Q)/L + K(Q)(Q)/(4L) + K(Q)(–Q)/(3L) + K(Q)(Q)/(2L) + K(Q)(–Q)/2L + K(–Q)(Q)/(5L) + K(–Q)(–Q)/(4L) + K(–Q)(Q)/(3L) + K(–Q)(–Q)/(2L) + K(–Q)(Q)/L
   	=	–(37/10)KQ2/L

   

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2. Two point charges of q₁ = 3.0 μC and q₂ = 4.0 μC are situated at the opposite corners of a rectangle as shown below. The short side has length L = 0.25 m. Find the total potential at points A and B. If a free particle of charge q_f = 1.0 μC and mass M = 15 g has speed v_A = 2.50 m/s at point A and it follows the indicated path, what will be its speed at point B?

   

   Electrostatic potential for point charges is given by

   .

   At point A, the potential is

   V_A = kq₁/L + kq₂/Ö2L = 1.798 × 10⁵ Volts .

   At point B, the potential is

   V_B = kq₁/Ö2L + kq₂/L = 1.9778 × 10⁵ Volts .

   So point B is at a higher potential than point A.

   Conservation of Energy relates speed (Kinetic Energy) to potential. Using this principle we have

   ½mv_A² + q_fV_A = ½mv_B² + q_fV_B .

   Solving for v_B yields

   v_B = { v_A² + 2q_f(V_A – V_B)/m }^½ = 1.96 m/s .

   This results makes sense as a positive charge should slow down as it reaches an area of higher potential (i.e. more positively charge area).

   

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3. Find the electrostatic potential difference between points A and B which are distances r_A = 2.0 m and r_B = 1.0 m from an infinitely long thin wire with λ = 1.0 μC/m. The result E = λ/2pε₀r is useful. If an electron (q = –e = –1.602 × 10^–19 C and mass m_e = 9.11 × 10^–31 kg) is released from rest at point A, what is it's speed at point B?

   The electrostatic potential difference between two points when the electric field is known is given by

   ΔV = V_B – V_A = –ò_A^B E·dL .

   We are given the form of the electric field. In the diagram below the electric field and one possible path is shown.

   

   The electric field will be E = r λ/2pε₀r in this coordinate system which has cylindrical symmetry. As well, the path element is dL = rdr + qrdθ + z sin(θ)dz where r, q, and z are the cylindrical coordinate unit vectors. The potential difference is thus

   DV = –ò₂¹ (l/2pε₀) dr/r = (l/2pε₀) ln(2) = 12465 Volts .

   The result indicates that point B is at a higher potential than point A which we expect since point B is closer to the positively charged wire.

   Conservation of Energy relates speed (Kinetic Energy) to potential. Using this principle we have

   ½mv_A² + q_fV_A = ½mv_B² + q_fV_B .

   Solving for v_B yields

   v_B = { v_A² + 2q_f(V_A – V_B)/m }^½ = {–2qΔV/m }^½ = 6.62 × 10⁷ m/s .

   This result makes sense as a negative charge should speed up as it approaches the positive wire.

   

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4. Find the electrostatic potential between points A and B which are distances r_A = 2.0 m and r_B = 1.0 m from an infinitely large thin plate with Σ = 1.0 μC/m² . The result E = Σ/2ε₀ is useful. If an electron (q = –e = –1.602 × 10^–19 C and mass m_e = 9.11 × 10^–31 kg) is released from rest at point A, what is it's speed at point B?

   The electrostatic potential difference between two points when the electric field is known is given by

   ΔV = V_B – V_A = –ò_A^B E·dL .

   We are given the form of the electric field. In the diagram below the electric field and one possible path is shown.

   

   The electric field will be E = i Σ/2ε₀ in this coordinate system which has Cartesian symmetry. The path element is dL = idx + jdy + kdz. The potential difference is thus

   DV = –ò₂¹ (s/2ε₀) dx = Σ/2ε₀ = 56497 Volts .

   The result indicates that point B is at a higher potential than point A which we expect since point B is closer to the positively charged wire.

   Conservation of Energy relates speed (Kinetic Energy) to potential. Using this principle we have

   ½mv_A² + q_fV_A = ½mv_B² + q_fV_B .

   Solving for v_B yields

   v_B = { v_A² + 2q_f(V_A – V_B)/m }^½ = {–2qΔV/m }^½ = 1.41 × 10⁸ m/s .

   This result makes sense as a negative charge should speed up as it approaches the positive plate.

   

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

   

   Find the energy stored in the field.

   The energy stored in the electric field is found by evaluating

   .

   Since the field is radial and piecewise, there are three contributions to the energy

   .

   Evaluating the integrals yields

   

   

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

   

   Find the energy stored in the field.

   The energy stored in the electric field is found by evaluating

   .

   Since the field is radial and piecewise, there are three contributions to the energy

   .

   Evaluating the integrals yields

   

   

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7. A thick conducting spherical shell has out radius R and thickness a. A charge 4Q is placed on the conductor. Then a charge –2Q is placed at the centre of the shell. Find and sketch the electric field as a function of r.

   

   First, remember that charge can only reside on the surface of a conductor in equilibrium. Placing a charge of –2Q at the centre of the sphere induces a charge of +2Q on the inner side of the shell. Since charge is conserved, a further charge of –2Q must be added to the outer shell. That means that the total charge on the outer shell is 2Q.

   For cases of spherical symmetry, Gauss' Law

   ò E • ndA = Q_enclosed/ε₀ ,

   becomes

   E(b) 4pb² = Q_enclosed/ε₀ .

   The inner and outer surfaces of the spherical shell forms two boundaries. We will need apply Gauss’ Law to three cases; one where the Gaussian surface is smaller than the first boundary, second between the two boundaries, and third where it is larger than the outer boundary.

   Case i. b < R – a

   We sketch a side view of the sphere and the Gaussian surface below.

   

   We know that the enclosed charge is the free charge Q_enclosed = –2Q. Thus Gauss’ Law yields,

   E(b) 4pb² = –2Q/ε₀ .

   Thus the electric field is E(a) = –2Q/4pε₀b².

   Case ii. R – a < b < R

   The Gaussian surface looks like

   

   Remember that there can be no free charge inside a conductor. Thus the total charge inside the Gaussian sphere consists of the free charge at the centre plus the induced charge on the inner surface of the spherical shell. Hence Q_enclosed = –2Q + 2Q = 0. Since the total charge is zero, the electric field must be zero as well. This makes sense since there cannot be any electric field inside a conductor at equilibrium.

   Case iii. b > R

   The Gaussian surface looks like

   

   The "Gaussian sphere" encircles the entire sphere. The total charge inside the Gaussian sphere consists of the free charge at the centre plus the induced charge on the inner and outer surfaces of the spherical shell. Hence Q_enclosed = –2Q + 2Q + 2Q = 2Q.

   Thus Gauss’ Law yields,

   E(b) 4pb² = 2Q/ε₀ .

   Thus the electric field is E(b) = 2Q/4pε₀b².

   The electric field as a function of a looks like

   

   

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8. A charge Q is placed near two very large conducting plates which meet at a right angle as shown in the diagram below. Use the method of image to determine the components of the force exerted on the charge by the wall. Determine the induced charge density s on each wall.

   

   The conducting plates have to be at zero potential. The arrangement of the original charge and any image charges must be such that V_plate = 0. For a single straight plate we know that there one be one image charge behind the plate of opposite sign. With two plates we might guess that we need two image charges as shown below.

   

   Note that the potential at the indicated point will not be zero since

   V_plate = KQ/R₁ + K(–Q)/R₁ + K(–Q)/R₂ ¹0 .

   It appears that we will need a third, positive, image charge for the net potential to be zero at any point on the plates. Symmetry demands that it be located at (–a,–b) relative to the corner.

   

   Now we have

   V_plate = KQ/R₁ + K(–Q)/R₁ + K(–Q)/R₂ + KQ/R₂ = 0

   as is required.

   Now the force on the real charge because of the induced charge on the plates is the same as that which would be produced by the image charges. The force diagram looks like

   

   The forces due to the individual images acting on the real charge are

   F₁ = k(Q)(–Q)/(2a)² ,

   F₂ = k(Q)(Q)/[(2a)² + (2b)²] ,

   F₃ = k(Q)(–Q)/(2b)² .

   The angle is q = arctan(b/a).

   The net force in the x direction is

   F_x = F₂cos(q) – F₁ = ¼KQ²[2a/(a²+b²)^3/2 – 1/a²] .

   The net force in the y direction is

   F_y = F₂sin(q) – F₃ = ¼KQ²[2b/(a²+b²)^3/2 – 1/b²] .

   The charge density s is determined by the requirement that s = ε₀E_charges, where E_charges is the net electric field due to the real and image charges. Recall that E_charges must be perpendicular to the surface of a conductor. If we label the corner as (0,0), then E_charges at (0,y) is

   

   The charge density is

   

   Following the same steps for a point (0,x) along the horizontal wall we find

   .

   There is one check that we can do to see if our answer is correct. The net induced charge on the plates must be –Q. Thus we need to check that the following identity is true

   ò_0¥ s (y) dy + ò_0¥ s (x) dx = –Q .

   And it does.

   

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9. A charge Q is placed a distance a from the centre of a conducting thin spherical shell of radius R (a < R). The shell is kept at zero potential. Find the location, sign, and magnitude of the image charge in this situation. What is the force exerted on the original charge by the shell? Determine the induced charge density s (q) on the shell.

   

   Symmetry would lead us to guess that we need a single charge q at (0,b), where we have to determine q and b, as shown below.

   

   The potential at the point D would be

   V_sphere = kQ/z + kq/r .

   Expressing z and r in terms of R, a, b, and q , we find

   V_sphere = kQ/[a² + R² – 2aRcos(q )]^½ + kq/[b² + R² – 2bRcos(q )]^½ .           (1)

   The easiest way to proceed is to evaluate V_sphere = 0 at two points, say A and B, find the value of q and b, and then show that the results work in general for any point. At point A we have

   kQ/(R–a) + kq/(b – R) = 0 .           (2)

   At B, we have

   kQ/(R+a) + kq/(b + R) = 0.           (3)

   Solving the equations (2) and (3) together, we find q = –(R/a)Q and b = R²/a. Next we substitute these results into equation (1)

   Vsphere	= kQ/[a2 + R2 – 2aRcos(q )]½ – k(R/a)Q/[(R2/a)2 + R2 – 2R(R2/a)cos(q )]½
   	= kQ/[a2 + R2 – 2aRcos(q )]½ – kRQ/[(R2)2 + a2R2 – 2aR3cos(q )]½
   	= kQ/[a2 + R2 – 2aRcos(q )]½ – kRQ/{R[R2 + a2 – 2aRcos(q )]½}
   	= kQ/[a2 + R2 – 2aRcos(q )]½ – kQ/[R2 + a2 – 2aRcos(q )]½
   	= 0

   So indeed the potential is zero everywhere on the surface of the sphere.

   The force on the real charge because of the induced charge on the shell is equal to the force on the charge due to the image charge

   F	= kQq/(b–a)2
   	= –kQ(R/a)Q/[R2/a–a]2
   	= –kQ2 [aR/(R2–a2)2]

   The minus sign indicates that the real charge is attracted to the image charge.

   The charge density s will be a function of the angle q . We know s = ε₀E_charges, where E_charges is the net electric field due to the real and image charges. Recall that E_charges must be perpendicular to the surface of a conductor.

   

   First note that z = i [Rcos(q ) – a] + j Rsin(q ) and that r = –i [b – Rcos(q )] + j Rsin(q ). Now b= R²/a . The net electric field is

   Enet =	EQ + Eq
   =	kQz/z3 + kqr/r3
   =	(KQ/z3){ z – (a/R)2 r }
   =	(KQ/z3){ i [Rcos(q ) – a + (a/R)2(b – Rcos(q ))] + j [Rsin(q ) – (a/R)2Rsin(q )] }
   =	(KQ/z3)R[1 – (a/R)2] { i cos(q ) + j sin(q ) }

   Note that electric field is radial since i cos(q ) + j sin(q ) is the unit radial vector

   

   where u is the unit radial vector. Thus the charge density is

   .

   Checking that this result is reasonable, we use MAPLE to integrate over the entire surface of the sphere and find

   Q_induced = R² ò₀^2p df ò₀^p s(q) sin(q) dq = –Q .

   This is right since we expect that the induced charge on the shell be equal and opposite to the charge at (0, a).

   

Questions? mike.coombes@kwantlen.ca