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6 * 10 ^-3 towards charge A
Force on charge C = kq^2 ( 1/rb^2+1/ra^2)
=28.50 *10 ^-3 N outwards

(2) An electric charge Q = 4.50 uC is in a region of electric field with a y-component Ey= 7000 N/C, an x component Ex =6000 N/C and a z-component Ez = 0.
What are the magnitude and direction of the force on the charge Q?
sol : E^2= Ex^2 + Ey^2 = 49,000000 + 36,000000
E= 9219.5 N
Direction = Ex/Ey =tan (theta) = 6000/7000 = .85
Theta = 40.1 degree
Force = qE = 4.50 *10^-6* 9219.5 N = .041 N at 40 degree to horizontal axis

(3)To move a charged particle through an electric potential difference of 2.0 x 10^3V requires 4 x 10^-6 J of energy. What is the magnitude of the charge?
Sol : P.D = E/C = 4* 10^ -6 /C
C= 2 nC

4)The places of a capacitor are separated by 0.18mm of air and each have a potential difference of 56v between them. If the area of each plate is 100cm^, what is:
(a) the capacitance,
(b) the electric field strength between the plates,
(c) the amount of charge stored on each plate?
Sol: C= Eo( A/D) = 8.85* 10^-12 * .01/56 = 1.5 *10 ^-15 C
Electric field = !/2 CV^-2 = 0.5 * 1.5 *10 ^-15 * 56*56 = 2.52 * 10^-13 V
Q= CV = 1.5 *10 ^-15 * 56 = 84 *10 ^-15 C

(5)Two 90nF capacitors are to be connected together. What would be the total capacitances of the pair if they where connected in series, and in parallel?
Sol : In parallel = 90+90 = 180 nF
In series = 1/90+1/90 = 2/90 .022 nF

(6)An electrolyric capacitor is rated at 76uF at 150v.
(a) How much charge can be stored in the capacitor?
(b) How much energy can it store?

Sol :
Charge = CV = 76 *150 = 11400* 10 ^-9 C
E= ½ QV^2 = 6.4 * 10^-11 J

(8) Two 2.8cm long coils of wire are wrapped around the same magnetic core, of relative permeability 72, giving them a coupling factor of 0.92. Coil A has 1880 turns and coil B has 1500 turns. Find:
(a) the self inductance of coil A,
(b) the self inductance of coil B,
(c) the mutual inductance between the two coils,
(d)the emf that would be generated in coil A if the current in coil A changes at a rate of 2.7 A/s,
(e) the emf that would be generated in coil B if the current in coil A chances at a rate of 2.7 A/s.
Sol : permeability = 72
Coupling factor = 0.92
L= 2.8 cm
N1= 1880
N2= 1500
Self inductance for A = n1A mu /l = 0.008
Self inductance for A = n2A mu /l = 0.010

Mutual inductance = k (L1L2) ^1/2 = 0.0082

EMf induced in coil A = R1L1+ L1 DI1/dt + MDI2/dt
= 2.8^2 /2*3.14* 1880 + .0082 * 2.7 = 0.2220

EMf induced in coil A = R2L2+ L1 D21/dt + MDI1/dt
=0.2223

(9)
Convert the following numbers
(a)From binary to decimal, 10101011 and 101100
(b)from decimal to binary 47, and 259
(c)from hexadecimal to binary, 63, and F3
(d)from decimal to hexadecimal, 319, and 35.