course Phy 232 7/27 at 7:00 p.m.
.............................................
Given Solution: ** IL is the source. The law is basically an inverse square law and the angle theta between IL and the vector r from the source to the point also has an effect so that the field is B = k ' I L / r^2 * sin(theta). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ` Self-critique Rating: OK ********************************************* Question: `qWhat would be the area of a .20 F capacitor if plates are separated by 2.2 mm of air? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: A = d*C*4*pi*k = (0.0022 m)*(0.20F)/(8.85*10^-12 F/m) = 4.97*10^7 m^2 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** If a parallel-plate capacitor holds charge Q and the plates are separated by air, which has dielectric constant very close to 1, then the electric field between the plates is E = 4 pi k sigma = 4 pi k Q / A, obtained by applying Gauss' Law to each plate. The voltage between the plates is therefore V = E * d, where d is the separation. The capacitance is C = Q / V = Q / (4 pi k Q / A * d) = A / (4 pi k d). Solving this formula C = A / (4 pi k d) for A for the area we get A = 4 pi k d * C If capacitance is C = .20 F and the plates are separated by 2.2 mm = .0022 m we therefore have A = 4 pi k d * C = 4 pi * 9 * 10^9 N m^2 / C^2 * .20 C / V * .0022 m = 5 * 10^7 N m^2 / C^2 * C / ( J / C) * m = 5 * 10^7 N m^2 / (N m) * m = 5 * 10^7 m^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `qquery univ 24.50 (25.36 10th edition). Parallel plates 16 cm square 4.7 mm apart connected to a 12 volt battery. What is the capacitance of this capacitor? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Capacitance is equal to the charge of the plates divided by the potential difference. The potential difference is equal to the electric field times the distance between the plates. C = epsilon*A/d = (8.85*10^12 F/m)(0.16 m)^2/0.0047 m = 4.82*10^-11 F The Charge each plate is equal to the product of the potential difference and capacitance of the plates. Q = V*C = (12 V)*(4.82 *10 ^-11 C/V) = 5.78*10^-10 C The Electric Field is equal to the potential difference between the plates divided by the charge of the plates: E = V/d = Q/(C*d) = (5.78*10^-10)/(4.82*10^-11 C/V*0.0047 m) = 2551 V/m The Energy of the capacitor is equal to the work of the capacitor W = V*q = (12 V)*5.78*10^-10 = 6.936*10^-9 J confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** Fundamental principles include the fact that the electric field is very neary constant between parallel plates, the voltage is equal to field * separation, electric field from a single plate is 2 pi k sigma, the work required to displace a charge is equal to charge * ave voltage, and capacitance is charge / voltage. Using these principles we reason out the problem as follows: If the 4.7 mm separation experiences a 12 V potential difference then the electric field is E = 12 V / (4.7 mm) = 12 V / (.0047 m) = 2550 V / m, approx. Since the electric field of a plane charge distribution with density sigma is 2 pi k sigma, and since the electric field is created by two plates with equal opposite charge density, the field of the capacitor is 4 pi k sigma. So we have 4 pi k sigma = 2250 V / m and sigma = 2250 V / m / (4 pi k) = 2250 V / m / (4 pi * 9 * 10^9 N m^2 / C^2) = 2.25 * 10^-8 C / m^2. The area of the plate is .0256 m^2 so the charge on a plate is .0256 m^2 * 2.25 * 10^-8 C / m^2 = 5.76 * 10^-10 C. The capacitance is C = Q / V = 5.67 * 10^-10 C / (12 V) = 4.7 * 10^-11 C / V = 4.7 * 10^-11 Farads. The energy stored in the capacitor is equal to the work required to move this charge from one plate to another, starting with an initially uncharged capacitor. The work to move a charge Q across an average potential difference Vave is Vave * Q. Since the voltage across the capacitor increases linearly with charge the average voltage is half the final voltage, so we have vAve = V / 2, with V = 12 V. So the energy is energy = vAve * Q = 12 V / 2 * (5.76 * 10^-10 C) = 3.4 * 10^-9 V / m * C. Since the unit V / m * C is the same as J / C * C = J, we see that the energy is 3.4 * 10^-9 J. Pulling the plates twice as far apart while maintaining the same voltage would cut the electric field in half (the voltage is the same but charge has to move twice as far). This would imply half the charge density, half the charge and therefore half the capacitance. Since we are moving only half the charge through the same average potential difference we use only 1/2 the energy. Note that the work to move charge `dq across the capacitor when the charge on the capacitor is `dq * V = `dq * (q / C), so to obtain the work required to charge the capacitor we integrate q / C with respect to q from q = 0 to q = Q, where Q is the final charge. The antiderivative is q^2 / ( 2 C ) so the definite integral is Q^2 / ( 2 C). This is the same result obtained using average voltage and charge, which yields V / 2 * Q = (Q / C) / 2 * Q = Q^2 / (2 C) Integration is necessary for cylindrical and spherical capacitors and other capacitors which are not in a parallel-plate configuration. ** query univ 24.51 (25.37 10th edition). Parallel plates 16 cm square 4.7 mm apart connected to a 12 volt battery. If the battery remains connected and plates are pulled to separation 9.4 mm then what are the capacitance, charge of each plate, electric field, energy stored? The potential difference between the plates is originally 12 volts. 12 volts over a 4.7 mm separation implies electric field = potential gradient = 12 V / (.0047 m) = 2500 J / m = 2500 N / C, approx.. The electric field is E = 4 pi k * sigma = 4 pi k * Q / A so we have Q = E * A / ( 4 pi k) = 2500 N / C * (.16 m)^2 / (4 * pi * 9 * 10^9 N m^2 / C^2) = 5.7 * 10^-7 N / C * m^2 / (N m^2 / C^2) = 5.7 * 10^-10 C, approx.. The energy stored is E = 1/2 Q V = 1/2 * 5.6 * 10^-10 C * 12 J / C = 3.36 * 10^-9 J. If the battery remains connected then if the plate separation is doubled the voltage will remain the same, while the potential gradient and hence the field will be halved. This will halve the charge on the plates, giving us half the capacitance. So we end up with a charge of about 2.8 * 10^-10 C, and a field of about 1250 N / C. The energy stored will also be halved, since V remains the same but Q is halved. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didnt take the average of the potential difference when I was finding the energy. The potential difference between the capacitor and the battery is 12 V. The potential difference between the capacitor and the capacitor is 0, So midway between the two the potential difference is 6 V. Self-critique Rating: ********************************************* Question: `qquery univ 24.68 (25.52 10th edition). Solid conducting sphere radius R charge Q. What is the electric-field energy density at distance r < R from the center of the sphere? What is the electric-field energy density at distance r > R from the center of the sphere? What is the total energy in the field of this sphere? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I know that inside the sphere, the electric field will be 0. So u = o for r< R Outside the sphere the electric field energy will equal: (k*q)^2*epsilon/(2*r^4) The density of the electric field energy is the total electric field energy at a distance from the radius, divided by the volume of the electric field at that distance. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** We will find the energy density function then integrate that density function over all of space to find the total energy of the distribution. We will compare this with the energy required to assemble the distribution, and will find that the two are equal. * To integrate the energy density over all space we will find the total energy in a thin spherical shell of radius r and thickness `dr, then use this result to obtain our integral. * Then we will integrate to find the work required to assemble the charge on the surface of the sphere. Energy density, defined by dividing the energy .5 C V^2 required to charge a parallel-plate capacitor by the volume occupied by its electric field, is * Energy density = .5 C V^2 / (volume) = .5 C V^2 / (d * A), where d is the separation of the plates and A the area of the plates. Since C = epsilon0 A / d and V = E * d this gives us .5 epsilon0 A / d * (E * d)^2 / (d * A) = .5 epsilon0 E^2 so that * Energy density = .5 epsilon0 E^2, or in terms of k = 4 pi / epsilon0 * Energy density = 1 / (8 pi k) E^2. Since your text uses epsilon0 I'll do the same on this problem. In this problem the epsilon0 notation makes a good deal of sense: For the charged sphere the electric field for r < R is zero, since there can be no electric field inside a conductor. For r > R the electric field is * E = Q / (4 pi epsilon0 r^2), and therefore * energy density = .5 epsilon0 E^2 = .5 epsilon0 Q^2 / (16 pi^2 epsilon0^2 r^4) = Q^2 / (32 pi^2 epsilon0 r^4). The energy density (i.e., the energy per unit of volume) between r and r + `dr is nearly constant if `dr is small. As we saw above the energy density will be approximately Q^2 / (32 pi^2 epsilon0 r^4). The volume of space between r and r + `dr is approximately A * `dr = 4 pi r^2 `dr. The expression for the energy lying in the shell between distance r and r + `dr is therefore approximately * energy in shell = energy density * volume = Q^2 / (32 pi^2 epsilon0 r^4) * 4 pi r^2 `dr = Q^2 / (8 pi epsilon0 r^2) `dr. This leads to a Riemann sum over radius r. As we let `dr approach zero the Riemann sum approaches an integral with integrand Q^2 / (8 pi epsilon0 r^2), integrated with respect to r. To get the energy between two radii we therefore integrate this expression between those two radii. If we integrate this expression from r = R to infinity we get the total energy of the field of the charged sphere. This integral gives us * total energy = Q^2 / (8 pi epsilon0 R) (alternatively using k = 4 pi k / epsilon0, this result would be k Q^2 / (2 R). This form is less complicated to the eye and will be used in the comparison below.) We compare this with the work required to charge the sphere: * To bring a charge `dq from infinity to a sphere containing charge q requires work k q / R `dq * To charge the entire sphere the work would therefore be the integral of k q / R with respect to q. * We integrate from q = 0 to q = Q, obtaining the total work required to charge the sphere. Our antiderivative is k (q^2 / 2) / r. If we evaluate this antiderivative at lower limit 0 and upper limit Q we get the total work, which is k Q^2 / (2 R). This agrees with our previous result, obtained by integrating the energy density of the field. Since k = 1 / (4 pi epsilon0) the work is k Q^2 / (2 R) = Q^2 / (8 pi epsilon0 R). So the energy in the field is equal to the work required to assemble the charge distribution. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating:"