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Electrostatic Forces and Energy Objective: To calculate electrostatic force, electric field vector, electric potential energy and electric potential for a given continuous or point charge distribution on a field point that may or may not contain a charge. Discussion: There are many relationships in static electricity that are of importance. Electrically charged particles exert forces on other charged particles by creating electric fields that alter their surroundings. These fields can do work on charged particles that can be measured for a given charge. The amount of energy a point source distribution can provide to a charged object can be determined specifically for that charged object (electric potential energy) or in general normalized by dividing by the amount of charge (electric potential). All of this information can be determined simply using Coulomb’s Law for electrostatic forces and the definition of electric field, electric potential energy and electric potential. Governing Equations: kq q FES 12 2 Eq0 1.1 r kq E 2 rˆ 1.2 r q PEE kq0 i 1.3 ri PEE V 1.4 q0 1.5 R Rx2 Ry2 Rz2 1.6 rˆiˆ Ex ˆ i E X circle r cos 1.7 rˆˆj Ey E ˆj rˆkˆ Ycircle r sin Ez ˆ k E Setting Up: Open Microsoft Excel® 1. Set up your spreadsheet exactly as you see below. A B C D E F G H I 1 K (Nm /c ) Major axis (m) Minor axis (m) 2 9e9 3 4 Source Points 5 Count Charge (Coul) X coordinate (m) Y coordinate (m) Z coordinate (m) Rx (N/C) Ry (N/C) Rz (N/C) Potential (v) 2 2 J K 1 2 3 4 5 Sum Rx 6 Sum Ry 7 Sum Rz 8 9 10 L M N O P j k Field Point Charge (Coul) X Coordinate (m) Y Coordinate (m) Z Coordinate (m) Direction Magnitude i Electrostatic Force (N) Electric Field (N/C) Electric Potential Energy (J) Potential (V) 2. The first cell will be used to count numbers in order to help write formulas to make continuous charge distributions. Enter 0 in cell A6, 0.01 in cell A7 and 0.02 in cell A8. Highlight these cells and drag down to cell A634 to count up to 6.28 which will help make a complete circle using polar coordinates on the XY coordinate plane. 3. Columns B, C, D, and E along with M2-5 will be data entry cells in which you will enter the source points and the field point information. Leave these cells empty for now. 4. In cells F6, G6, and H6 you will calculate the X, Y, and Z components of the electric field. To do this type “=IF(B6=0,0,$A$2*B6/((C6-$M$3)^2+(D6$M$4)^2+(E6-$M$5)^2)^(3/2)*($M$3-C6))” into cell F6 and “=IF(B6=0,0,$A$2*B6/((C6-$M$3)^2+(D6-$M$4)^2+(E6$M$5)^2)^(3/2)*($M$4-D6))” into cell G6. For the Z component type “=IF(B6=0,0,$A$2*B6/((C6-$M$3)^2+(D6-$M$4)^2+(E6$M$5)^2)^(3/2)*($M$5-E6))” into cell H6. The first IF function is required so that the unused points will not be divided by empty cells (zero). Now drag each formula down 630 cells so that your spreadsheet can accommodate up to 630 three dimensional source points. 5. The electric potential for each point can be calculated in column I using Equation 1.3 and 1.4 specifically for each point charge. Enter “=IF(B6=0,0,$A$2*B6/((C6-$M$3)^2+(D6-$M$4)^2+(E6-$M$5)^2))” into cell I6. Note that this is cell I6 not the number 16. Again the IF function is used to prevent from dividing by zero. Double click the formula box. 6. The sum of the X and Y components of the electric field vector can be determined in cells H5 and H6. Enter “=SUM(F6:F636)” into cell K5, “=SUM(G6:G636)” into cell K6, and “=SUM(H6:H636)” into K7. 7. The magnitude of the electric field vector can be calculated in cell J8 using Equation 1.5. Enter “=SQRT(K5^2+K6^2+K7^2)” into cell M8. The direction of the electric field vector depends on the direction of the electrostatic force on a positive test charge at that location, so that will be calculated after the direction of the force. 8. The magnitude of the electrostatic force can be calculated in cell M7 using the following formula from Equation 1.1: “=M2*M8”. After typing the formula place the curser between the M and the 2 in the formula and press the F4 key on the keyboard. This is a quick way to switch it to a fixed reference. 9. Since electric potential energy and electric potential are scalar quantities, they do not require a direction. Enter the following formula into cell M9 to calculate electric potential energy: “=M2*SUM(I6:I636)”. Note that this is cell I6 not the number 16 10. Determining direction: The unit vector with the appropriate direction can be determined using the resultants calculated in cells K5-7. The X, Y and Z components of the unit vector can be calculated in Columns N, O and P respectively. Type “=$K$5/SQRT($K$5^2+$K$6^2+$K$7^2)” in cell N8, “=$K$6/SQRT($K$5^2+$K$6^2+$K$7^2)” in cell O8 and “=$K$7/SQRT($K$5^2+$K$6^2+$K$7^2)” in cell P8. The electrostatic force can be in the direction of the field if the field point is positive and opposite the direction of the field if the field point is negative. Write an if function in cells N7-P7 to account for this. Enter “=IF($M$2<0,-N8,N8)” into cell N7 and drag it to the right. 11. Calculate electric potential by summing the potentials of each point charge in cell M10 by typing the following formula: “=SUM(I6:I636)” 12. Make a graph of all the X and Y coordinates using an XY scatterplot including up to row 634. This will display your charge distribution placed in the XY Plane. 13. To start place a single point charge of 1.0 microcouloumbs on (1,1,1) Development Questions: 1. Why is the IF function used in the formulas for Rx, Ry, Rz and Potential? 2. What do the following functions mean? A. =SQRT() B. =SUM() C. ATAN() D. PI(). 3. Quick test: What should be the magnitude of the direction vector (ijk) for the force and electric field? Write a formula in cells Q7 and Q8 to test and record the formula and its result below. 2 Dimensional Questions: Answer questions on a separate sheet of paper. 1. Place a charge of 1 microcoulomb (Hint: type “1e-6” to simplify) on 0,0 as a source point and then place a charge on the field point of 1 microcoulomb on (1,0,0). A. Predict the field and the force for a field point at (2,0) then test. B. Do the same for (3,0). C. What happens to electrostatic force and electric field if the charge of the source point is doubled? D. What if the charge of the field point is doubled? E. What if the field point has an opposite charge? F. What if the source point has an opposite charge? 2. Two point charges of 10.0 microcoulombs and -5.0 microcoulombs are located at (1.5,0,0) m and (-1,0,0) m respectively. What location (on the X axis) should a 1 microcoulomb charge be placed in order to have zero net force acting on it? Use goal seek and setting Electric Field to 0 by changing the X coordinate of the field point (the Y coordinate should be zero). Hint: use -2 as a starting value for goalseek. 3. Two spheres that can be treated as point charges are separated by a distance of 0.0500 m and have accumulated opposite charges. The electric field halfway between them is 30,000 N/C. Calculate the charge on each sphere. Hint: Make the charge on the second sphere equal to that of the first using a formula and then use goal seek. 4. Four point charges with charges of 3, 4, 5, and -6 microcoulombs are located at (0,1), (1,0), (-1,0) and (0,-1) respectively. What is the electric field magnitude, electrostatic force, electric potential energy and voltage at the origin? The field point has a charge of 1.00 microcoulombs. 2 Dimensional AP Questions: 1. Electric potential energy is path independent. If two positive point charges of 4.0 microcoulombs are present at (1,0) and (0,0), how much kinetic energy is gained by a positive point charge of 2.0 microcoulombs that is placed at rest and released at (2,0) and allowed to accelerate to (5,0)? What is the mass of the point charge if it has a velocity of 5.0 m/sec when it reaches (5,0)? 2. Two positive point charges of 100 microcoulombs are on the X axis at a position of (1,0) and (0,0) and there are two negative 100 microcoulomb point charges placed at (1,5) and (0,5). What is the kinetic energy of a particle with a charge of 10 microcoulombs that is released from rest at point (0.5,0) as it passes point (0.5,5)? Calculate the mass that the charge must have to attain a velocity of 50.0 m/sec over that interval. If a proton underwent the same displacement, what would its velocity be? (look up the mass and charge of an electron and proton) 3. A proton is fired directly at a uranium nucleus at 8.0*107 m/sec . The nucleus is large enough to be considered at rest throughout the problem. How close to the U nucleus does the proton reach if the atomic number of the nucleus is 92. Assume the effects of the electrons are negligible. Use manual guess and check since goal seek does not like very small numbers. (your answer should be given in nanometers) Continuous Charge Distributions: The superposition principle can be used to analyze continuous charge distributions. This can be done by breaking a continuous charge up into a large number of point charges very close together. To do this choose a large number of points (more than 100) and space them uniformly using a formula for the charge and the X and Y coordinates. See the answer key for specific instructions. 4. Using a 20.0 meter long line of charge along the Y axis centered on the origin with charge per unit length 1.5*10-10 C/m, find the location on the X axis at which the electric field is 2.5 N/C. Assume the line of charge can be replaced with 400 point charges This is a continuous charge distribution. Therefore the distribution has to be broken up into a large number of point charges spaced close together compared to the distance from the distribution. In order to do this we will break up the line of charge into 400 small increments along the Y-axis centered at the origin. To do this divide total charge (1.5*10-10*20) by 400 and place that result in cells B6 through B406. 3 Dimensional Questions: Setting Up: In order to place a ring of charge polar coordinates can be used. Column A counts the angle in radians that can be used and Equations 1.7 can be used to determine the X and Y coordinates of the point charges that will create the continuous charge distribution. Use the following formula “=$C$2*COS(A6)” in cell C6 and “=$C$2*SIN(A6)” in cell D6 to create the circle. Double click the formulas and a ring of 628 source point coordinates will be placed in a circle with the radius specified in cell C2. The charge of each point must be 1/628th of the total charge of the ring. This can be done simply by dividing the charge by 628 and entering that constant number repeatedly. For the ellipse refer to the major and minor axis for the X and Y coordinate formulas respectively. 1. Place a uniformly distributed ring of charge with a radius of 0.25 meters in the XY plane centered at the origin. The total charge on the ring is 35.0 nc. What is the magnitude and direction electric field at each of the following locations: A. (0,0,0.5) m B. (0,0,-0.5) m C. (0,0,1.0) m D. (5,0,0) m E. (0,5,0) m F. (0,0,5) m G. (0,0,0) m 2. Replace the ring of charge with an ellipse with a major axis equal to 0.05 m and a minor axis of 0.025 m. Repeat the same data points A-G from above. 3. Now replace the ring of charge with a single point charge of 35 nc at the origin. Repeat data points A-F. 4. What must be the radius of a circular charge distribution of 0.45 c in order to have a resulting electric field of 1000. N/C at (0,0,0.75)? Use goal seek. 5. A ring shaped conductor with radius a = 2.50 cm has total positive charge Q = 0.125 nC. At what distance along the central axis of the ring does the electric field vector have a magnitude of 7.0 N/C. Use goal seek.