Electrostatics Solving Problems folk.uio.no/ravi/emt013/ 1
Coulomb's Torsion Balance This dial allows you to adjust and measure the torque in the fibre and thus the force restraining the charge This scale allows you to read the separation of the charges
Experiments Results F Line Fr r 3
Experiments show that an electric force has the following properties: (1) The force is inversely proportional to the square of separation, r, between the two charged particles. 1 F () The force is proportional to the product of charge q1 and the charge q on the particles. r F q q 1 4
(3) The force is attractive if the charges are of opposite sign and repulsive if the charges have the same sign. F q 1 q r 5
Coulomb s Law The electrostatic force of a charged particle exerts on another is proportional to the product of the charges and inversely proportional to the square of the distance between them. F K q 1 r q 6
F K q 1 q r where K is the coulomb constant = 9 10 9 N.m /C. The above equation is called Coulomb s law, which is used to calculate the force between electric charges. In that equation F is measured in Newton (N), q is measured in unit of coulomb (C) and r in meter (m). 7
Permittivity constant of free space The constant K can be written as K 1 4 where is known as the Permittivity constant of free space. = 8.85 x 10-1 C /N.m K 1 1 4 4 8.8585 10 1 9 10 9 N. m / C 8
Example 1 Calculate the value of two equal charges if they repel one another with a force of 0.1N when situated 50cm apart in a vacuum. Solution q q F K 0.1 9 10 9 1 r (0.5) q q = 1.7x10 6 C = 1.7C 9
Example One charge of.0 C is 1.5m away from a 3.0 C charge. Determine the force they exert on each other. 10
Example 3 The following three charges are arranged as shown. Determine the net force acting on the charge on the far right (q3 = charge 3). 11
Step 1: Calculate the force that charge 1 exerts on charge 3... It does NOT matter that there is another charge in between these two ignore it! It will not effect the calculations that we are doing for these two. Notice that the total distance dsta between charge 1 and 3 is 3.1 m, since we need to add 1.4 m and 1.7 m. 1
The negative sign just tells us the charges are opposite, so the force is attractive. Charge 1 is pulling charge 3 to the left, and vice versa. Do not automatically treat a negative answer as meaning to the left in this formula!!! Since all I care about is what is happening to charge 3, all I really need to know from this is that charge 3 feels a pull towards the left of 4.9e- N. 13
Step : Cl Calculate lt the force that thtcharge exerts on charge 3... Same thing as above, only now we are dealing with two negative charges, so the force will be repulsive. The positive sign tells you that the charges are either both negative or both positive, sotheforceisrepulsive. I know that charge is pushing charge 3 to the right with a force of.5e 1 N. Step 3: Add you values to find the net force. 14
Multiple l Charges in Dimensionsi F41 F1 Q Q1 + F Q3 13 Principle of superposition p + Q4 Force on charge is vector sum of forces from all charges F 1 F 1 F 1 3 F 1 4 15
Example 4 Two equal positive charges q=x10-6 C interact with a third charge Q=4x10-6 C. Find the magnitude and direction of the resultant force on Q. 16
F Qq qq (4 10 )( 10 ) 1 K 9 r (0.5) 6 6 9 9 10 0. N F Qq 0.4 F x F cos 0.9 0. 3 N 0.5 0.3 F y F sin 0.9 0. 17 0.5 N F x 0.3 0. 46 N F x y 0 17
Example 5 In figure what is the q - q resultant force on the 3 charge in the lower left corner of the square? F 13 Assume that q=110 7 C 1 4 and a = 5cm q F 14 - q F 1 18
F F F F 14 13 1 1 F F F F 1 a qq K F a 13 qq K F 3 a q q K F 14 a K F 19
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Equilibrium Example Two fixed charges, 1C and -3C are separated by 10cm as shown in figure below (a) where may a third charge be located so that no force acts on it?
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Example Two charges are located on the positive x-axis of a coordinate system, as shown in figure below. Charge q1=nc is cm from the origin, and charge q=-3nc is 4cm from the origin. What is the total force exerted by these two charges on a charge q3=5nc located at the origin? 4
N F 4 9 9 9 1 3 10 0.56 (0 04) ) 10 )(5 10 )( 10 9 ( 31 (0.04) N F 4 9 9 9 10 3 37 ) 10 )(5 10 )(3 10 9 ( N F 3 10 3.37 (0.0) F F F N F F F F 4 4 4 3 3 31 3 10.81 10 3.37 10 0.56 5
In figure shown, locate the point at -5q q which the electric field is zero? - + Assume a = 50cm a -5qq q V S E P 1 1 - + a d a+d E E 1 = E 1 4 q (0.5 d) 1 4 5 q ( d) d = 30cm 6
We have q 1 =10 nc at the origin, q = 15 nc at x=4 m. What is E at y=3 m and x=0 y P 3 q 1 =10 nc 4 x q =15 nc Find x and y components of electric field due to both charges and add them up 7
Recall lle =kq/r k/ and k=8.99 x 10 9 N.m /C Field due to q 1 E = 10 10 N.m /C 10 X10 9 C/(3m) = 11 N/C in the y direction. E y = 11 N/C E x = 0 y E 3 q 1 =10 nc 5 4 x q =15 nc Field due to q E = 10 10 N.m /C 15 X10 9 C/(5m) = 6 N/C at some angle φ Resolve into x and y components E y =E sin f = 6 * 3/5 =18/5 = 3.6 N/C E x =E cos f = 6 * ( 4)/5 = 4/5 = 4.84 8 N/C E y = 11 + 3.6 = 14.6 N/C E x = 4.84 8 N/C Magnitude E E x E y 8
E E y = 11 + 3.6 = 14.6 N/C E x = -4.8 N/C 3 q 1 =10 nc 4 x q =15 nc Magnitude of electric field E E x E y E 14.6 4.8 15.4N /C Using unit vector notation we can also write the electric field vector as: E 4.8i 14.6 j φ 1 = tan 1 E y /E x = tan 1 (14.6/ 4.8)= 7.8 deg 9
+q 1 +q What is the electric field in the lower left corner of the square as shown in figure? P 3 Assume that q = 1x10-7 C and a = 5cm. -q E E p E E E +q 1 +q 1 E3 E E 1 E 3 1 q 4 a 1 q 4 a 1 q 4 a E x P E 3 3 E y -q E E 1 30
Evaluate the value of fee 1, E, & E 3 E 1 = 3.6x10 5 N/C, E = 1.8 x 10 5 N/C, E 3 = 7. x 10 5 N/C We find the vector E need analysis to two components E x = E cos45 E y = E sin45 E x = E 3 E cos45 = 7.x10 5 1.8x10 5 cos45 = 6x10 5 N/C E y = E 1 E sin45 = 3.6x10 5 1.8 x10 5 sin45 = 4.8x10 5 N/C E E x E y = 7.7 x10 5 N/C tan 1 Ey Ex = 38.6 o 31
Example A particle having a charge q=310-9 C moves from point a to point b along a straight line, a total distance d=0.5m. The electric field is uniform along this line, in the direction from a to b, with magnitude E=00N/C. Determine the force on q, the work done on it by the electric field, and the potential difference Va-Vb. 3
The force is in the same direction as the electric field since the charge is positive; the magnitude of the force is given by F =qe = 310-9 00 = 60010-9 N The work done by this force is W =Fd = 60010-9 0.5 = 30010-9 J The potential difference is the work per unit charge, which is Va-Vb= W/q = 100V Or Va-Vb = Ed = 00 0.5 = 100V 33
Electric flux. (a) () Calculate the electric flux through the rectangle in the figure (a). The rectangleis10cmby0cmandtheelectric field is uniform with magnitude 00N/C. (b) What is the flux in figure if the angle is 30 degrees? The electric ti flux is EA E EAcos So when (a) =0, we obtain E EAcos EA 00 N/ C 0.10.m 4.0 N m C And when (b) =30 degrees, we obtain E EAcos30 00 N/ C 0.1 010. 0m cos30 3.5N 35Nm C 34
To calculate the electric flux due to a point charge we consider an imaginary closed spherical surface with the point charge in the center, this surface is called gaussian surface. Then the flux is given by E. da = E dacos ( = 0) = q r 4 da = q 4 r 4 r = q Note that the netflux through h aspherical gaussian surface is proportional to the charge q inside the surface. 35
Consider several closed surfaces as shown in figure surrounding a charge Q as in the figure below. The flux that passes through surfaces S 1, S and S 3 all has a value q/. Therefore we conclude that the net flux through any closed surface is independent of the shape of the surface. Consider a point charge located outside a closed surface as shown in figure. We can see that the number of electric field lines entering the surface equal the number leaving the surface. Therefore the net electric flux in this case is zero, because the surface surrounds no electric charge. 36
In figure two equal and opposite charges of Q and -Q what is the flux for the surfaces S 1, S, S 3 and S 4. Solution For S 1 the flux = zero Q For S the flux = zero S For S 3 the flux = +Q/ o S 1 For S 4 the flux = Q/ o S 3 S -Q S 4 37
Example What must the magnitude of an isolated positive charge be for the electric potential at 10 cm from the charge to be +100V? V 1 4 q r 1 q V 4 r 100 4 8.9 10 0.1 1.11 10 9 C 38
Example Aa Qq 1 Qq Qq 4 Aa P Aa q Aa Qq 3 What is the potential at the center of the square shown in figure? Assume that q1= 1 +110-8 C, q= -10-8 C, q3=+310-8 C, q4=+104-8 C, and a = 1m. 39
Solution Aa Qq 1 Qq Aa q P Aa V n V n 1 q 1 q q3 4 r The distance r for each charge from P is 0.71m q 4 Qq 4 Aa Qq 3 9 8 910 (1 3 ) 10 V 500V 0.71 40
V B -V A = W AB / q o V B V A W AB Ed q o V k q r 41
Example Two charges of µc and -6µC are located at positions (0,0) m and (0,3) m, respectively. (i) Find the total electric potential due to these charges at point (4,0) 0)m. (ii) How much work is required to bring a 3µC charge from infinity to the point P? 4
-6 (0,3) + (0,0) (4,0) P 43
Vp = V1 + V V q 1 q k r 1 r V 10 6 10 4 5 6 6 9 9 10 6.3 10 3 volt 44
(ii) the work required is given by W = q3 Vp = 3 10-6 -6.3 10 3 =-18.9 10-3 J The -ve sign means that work is done by the charge for the movement from to P. 45
Electric Potential Energy The definition of the electric potential energy of a system of charges is the work required to bring them from infinity to that configuration. q 1 q r 46
To workout the electric potential energy for a system of charges, assume a charge q at infinity and at rest as shown in figure. If q is moved from infinity to a distance r from another charge q1, then the work required is given by W=Vq V k q r 1 q 1 q q 1 q r U U W k k q1 q r r 1
To calculate l the potential energy for systems containing more than two charges we compute the potential energy for every pair of charges separately and to add the results algebraically. U k q i q j r ij 48
Example -4q Three charges are held fixed as shown in figure. What is the potential Aa energy? Assume that q=110-7 Canda=10cm a 10cm. Aa +1q Aa +q 49
-4q U=U1+U13+U3 Aa Aa U ( q)( 4q) k a U ( q)( q) a 10q k a +1q ( 4q)( q) a 9 7 10 (10)(1 10 ) U 9 10 0.1 9 3 J Aa +q 50
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Example Point charge of +110-9 C and -110-9 C are placed 10cm part as shown in figure. Compute the potential at point a, b, and c. Compute the potential energy of a point charge +410-9 C if it placed at points a, b, and c. Ac V V n n k q r i i Ab 10cm Aa 10cm +q 1 4cm 6cm 4cm +q 5
Ac At point a 10cm 10cm Ab Aa +q 1 4cm 6cm 4cm +q 9 9 9110 110 V a 910 900V 0.06 0.04 53
At point b Ac 10cm 10cm Ab Aa 4cm 6cm 4cm +q 1 +q 9 9 9110 110 V b 9 10 1930V 0.04 0.14 54
At point c Ac 10cm 10cm Ab Aa +q 1 4cm 6cm 4cm +q 9 9 9 1 10 1 10 V c 910 0V 0.1 0.14 55
We need to use the following equation at each point to calculate the potential energy, U = qv At point a Ua = qva = 410-9(-900) 900) =-3610-7J At point b Ub = qvb = 410-91930 = +7710-7J7J At point c Uc = qvc= 410-90 = 0 56
V B -V A = W AB / q o V B V A W AB Ed q q o V k r U k q rq 1 r 57
Example In the rectangle shown in figure, q1 = -5x10-6 C and q = x10-6 C calculate the work required to move a charge q3 = 3x10-6 C from B to A along the diagonal of the rectangle. 58
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Course Text Book Physicsfor scientists and engineering with modern physics. By R. A. Serway, Other Recommended Resources: Borowitz and Beiser Essentials of physics.. Addison Wesley Publishing Co., 1971. Halliday, D. and Resnick, R. Physics (part two). John Wiley & Sons, Inc., 1978. Kubala, T.S., Electricity : Devices, Circuits and Materials, 001 Nelkon, M. and Parker, P. Advanced level physics. Heinemann Educational Books Ltd., 198. Ryan, C.W., Basic Electricity : A Self Teaching Guide, 1986 Sears, F.W., Zemansky, M.W. and Young, H.D. University physics Addison Wesley Publishing Co., 198. Weidner, R.T. and Sells, R.L. Elementary physics: classical and modern. Allyn and Bacon, Inc., 1973. Valkenburgh, N.V., Basic Electricity: Complete Course,, 1993 60