Test 1 1. a) Write the definition of the electric field vector. b) Which attributes of particles determine their electromagnetic interaction? c) Describe, in terms of electrostatic force, the electrostatic interaction between charged particles. d) Draw the lines of an electric field produced by a dipole. 2. An electron1 enters the region of a uniform electric field produced by two parallel plates of length L = 10cm as shown in figure 23.25. The strength of the field is E = 200 N/m and the initial speed of the electron is vi = 3⋅106 m/s. a) Find the acceleration of the electron while it is in the electric field. b) Find the time is takes the electron to travel through the field. c) Determine the deflection angle. 3. Three “point charges” are located at the corners of an equilateral triangle with 0.5 m long side, as shown in Figure P23.7. Calculate2 the net electric force on the 7 µC “charge”. 4. A nonuniform electric field is given by the expression E(x,y,z) = [ay,bz,cx], where a,b and c are constants. Determine the electric flux through a rectangular surface in the xy-plane, extending from x = 0 to x = w and from y = 0 to y = h 5. A thin rod of length L and uniform linear charge density λ lies along the x-axis, as shown in Figure P23.35. Show that the electric field strength at point P, a distance y0 from the rod along the perpendicular bisector is given by E = 2ksinθ0/y0. 1 2 electron: charge -1.6⋅10-19 C; mass 9.11⋅10-31 kg Coulomb constant: 9⋅109 Nm2 /C2 -1- r r r a)The electric field vector at a certain position r is a vector E(r ) such that the electric force exerted on "a charge" q (a particle with charge q) placed at this position would be r r Fel = qE b) The electric charge and spin determine electromagnetic interactions. r c) The electrostatic force F21 , that particle 1 exerts on particle 2 depends on the charges, q1 and q 2 , of the particles, and the relative position of the particles. r kq q F21 = 12 2 rˆ12 r d) + - -2- a) From the definition of the electric field vector the force exerted on the electron is F = −eE = −eEˆj We can assume that only this electrostatic force is exerted on the electron, therefore from Newton’s second law the acceleration of the electron is ( L -------------------------vi E y +++++++++++++++++++ vf x ) F eE ˆ − 1. 6 ⋅10 −19 C ⋅ 200 N C ˆ a = = − j= j = −3. 5 ⋅ 1013 m 2 ˆj − 31 s m m 9.1⋅ 10 kg ( ) b) Since the acceleration of the electron is in the vertical direction, the x-component of the position is a linear function of time. The time required by the electron to pass the electric field is therefore ∆t = L L 0.1m = = = 3. 33 ⋅10 −8 s 6m v x v i 3 ⋅ 10 s c) In a motion with constant acceleration, velocity is a linear function of time. With the time reference t0 = 0s at the instant when the electron enters the field, velocity is described by the following funcion v = v i + at = [v i , 0] + [0, a ]t = [v i , at ] Using the scalar components of the velocity at the instant when the electron leaves the field, one can find the deflection angle 3.5 ⋅1013 m s 2 ⋅ 3.33 ⋅ 10− 8 s at θ = arctan = arctan ≈ 21° vi 3 ⋅ 106 m s θ -3F31 y q3 = 7µC F F32 c = 0.5m x q2 = -4µC q1 = 2µC µC particle by the other two particles. The electric field produced by the 2µC particle, at the location of the 7µC particle, multiplied by the charge of the latter gives the first force kq1q 3 [cos 60°, sin 60°] = c2 9 ⋅ 109 Nm2 ⋅ 2 ⋅ 10− 6 C ⋅ 7 ⋅ 10− 6 C F31 = q 3E1 = = ( C )( )( (0.5m ) 2 ) [cos 60°, sin 60°] = [0.252,0.436]N Similarly, the second force exerted by the -4µC particle is kq 2q 3 [− cos 60°, sin 60°] = 2 c 9 ⋅ 109 Nm2 ⋅ − 4 ⋅ 10− 6 C ⋅ 7 ⋅ 10− 6 C F32 = q 3E 2 = ( = C )( )( (0.5m ) 2 ) [− cos 60°, sin 60°] = [0.504,−0.872]N The net electric force on the 7µC is therefore F = F31 + F32 = [0.752,−0.436]N -4y h E y dy dx dA x x w z The orientation of the considered surface places all the differential surface elements in the xy-plane (the z-component for all elements has a zero value). Additionally, all vectors, assigned to these elements, are in the z-direction dA = [0,0, dA ] = [0, 0, dxdy ] The differential flux (through the marked element) is therefore dΦ = E ⋅ dA = [ay,0, c x] ⋅ [0,0, dxdy ] = c xdxdy Integration over the surface requires that the limits for x are 0 and w, while the limits for y are 0 and h. x2 Φ = ∫ ∫ cxdxdy = c ∫ 0 0 0 2 hw h w 0 w2 h cw 2 h dy = c ∫ dy = 2 0 2 -5- The electric field of an extended object requires integration of the dE y electric field produced by the P individual differential fragments of θ0 θ the object. For a linear object the charge dq of its fragment of length r y0 dx depends on the linear charge density λ. dq = λdx x O dx This statement defines the charge L density. A differential fragment can be considered as a particle (“point charge”). From Coulomb’s law, the electric field at point P, produced by the (marked) differential piece of the rod is therefore dq λdx λdx dE = k 2 rˆ = k 2 [ − sin θ , cos θ ] = k [− x, y 0 ] 3 2 2 2 r x + y 20 x +y ( 0 ) Integration of the above expression over the rod yields the electric field at point P. From the symmetry, one should conclude that only the y-component of the field is different than zero. L L L 2 2 λydx x 2kλ sin θ0 2 Ey = ∫ k = k λ y = 2 k λ y = 0 0 2 y0 −L y 20 x 2 + y20 −L (x 2 + y 20 )3 2 2 L 2 2 y 0 + y 20 2 Since the other components have zero values, the electric field strength is 2kλ sin θ 0 E = E 2x + E 2y + E 2z = E y = y0