RI Y6 Remedial E-field Soln

Raffles Institution Year 5-6 Physics Department 1 2024 Year 6 H2 Physics Remedial Chapter 13: Electric Fields Suggested Solutions 1 (a) EA = π − −− × ×× 19 12 3 2 4.8 10 4 (8.85 10 )(3.00 10 ) = 4.796 x 10–4 N C–1 EB = π − −− × ×× 19 12 3 2 1.60 10 4 (8.85 10 )(4.00 10 ) = 8.992 x 10–5 N C–1 Resultant electric field = −−×+× 42 52(4.796 10 ) (8.992 10 ) = 4.88 × 10–4 N C–1 (b) VA = πε − − −× × 19 3 ( 4.8 10 ) 4 (3.00 10 )o = –1.439 × 10–6 V VB = πε − − × × 19 3 (1.6 10 ) 4 (4.00 10 )o = 3.597 × 10–7 V V = VA + VB = –1.439 × 10–6 + 3.597 × 10–7 = –1.08 × 10–6 V 2 (a) The electric field strength at a point is defined as the electric force exerted per unit positive charge placed at that point. (b) Main features of drawing: 1. Arrows point from positive charge X to negative charge Y. 2. More field lines (nearly double) point towards the negative charge compared to the positive charge. 3. Asymmetry of the field lines between the two charges, with the ‘hump’ closer to the positive charge. Symmetry about the horizontal line joining the two charges. (c) (i) Distance TY = 220.20 0.10 0.03 0.173 cm−= = Force on T due to charge at Y, FY = ( )( ) ( ) 10 10 4 22 2 6 2 10 1 2 10 2 23 10 N4r 4 0 03 1 0 YT o o ..qq . .πε πε −− − − ×× = = × × 0.20 cm X Y

Raffles Institution Year 5-6 Physics Department 2 Force on T due to charge at X, FX = ( )( ) ( ) 10 10 4 22 2 3 2 10 1 2 10 3 45 10 N4r 4 0 10 1 0 XT o o ..qq . .πε πε −− − − ×× = = × × Magnitude of the net force acting on T, F = 2 24 4( 3.45 2.23 ) 10 4.11 10 N−−+ ×=× (ii) Direction of the net force on T is acting towards the top right hand corner. 3 (a) (i) The electric field strength at a point is defined as the electric force exerted per unit positive charge placed at that point. (ii) 1. Electric field strength inside both spheres is zero 2. There is a point between the spheres, where electric field strength is zero and the electric field strength of sphere A is positive at its surface (which means that the direction of the electric field is away from sphere A). (iii) The particle will move towards sphere B and oscillate about x = 8.0 cm 4 (a) (i) The electric field strength at a point is defined as the electric force exerted per unit positive charge placed at that point. (ii) ( ) 41 2 0 600 18750 1 88 10 N C3 20 10 VE .d. − − −−∆= = = = ×∆× (iii) ( )( ) 19 15 15 2 31 1 60 10 18750 3 293 10 3 29 10 m s9 11 10 net EFF ma qE .qEa ..m. − − − = = × = = = ×= ×× Assuming constant acceleration and using kinematics equation in the horizontal direction, ( ) 2 2 9 15 2 3 20 10 4 41 10 s3 293 10 1 2 2 s . . at t . ut s a − − = + = = × = ×× T FResultant FX FY X Y

Raffles Institution Year 5-6 Physics Department 3 5 (c) (i) Shows arrows from + to – plates on all the given lines (ii) Surface of constant potential. OR No work done in moving a charge on the surface. OR All points on the surface have the same potential. OR No potential difference between any two points on the surface. (iii) 3 vertical lines between plates, equally spaced. 0 V line is right in the middle between the plates, +12.5 kV line at ¼ way from the left. −12.5 kV line at ¾ way from the left. (iv) 1. Electric force, F = qE where E is the electric field strength given by E = −dV/dx Taking the magnitude of the electric force, ( ) ( ) 33 9 25 10 25 10 5.5 10 4.0 0.0688 mN 0.07 mN (shown) dVF qE q dx − = = − × −− × = × = = 2. Since electric force F is the only force acting on the particle, by Newton’s 2 nd Law Fnet = F = ma where a is the acceleration and m is the mass of the particle. 3 2 4 0.0688 10 0.573 m s1.2 10 Fa m − − − ×= = =× In the horizontal direction: Using kinematics equation, since a is constant, s = ut + ½ at2 The initial speed in the horizontal direction is zero. Hence u = 0. s = 0.t + ½ a t2 2.0 = ½ (0.573)t2 t = 2.64 s In the vertical direction: Length of plate = vertical constant speed x time = 0.80 x 2.64 = 2.11 m 6 (a) The electric field strength at a point is defined as the electric force exerted per unit positive charge placed at that point. (b) (i) Positive

Raffles Institution Year 5-6 Physics Department 4 (ii) ( ) ( ) ( ) ( ) 9 1 3 0 41 10 5000 0.237 N m 18 10 0.048 F kx qE qE qVk x xd − − − = ⇒= × ⇒= = = = × ∑ (c) (i) The electrons/charges flow from one plate to the other until there is no potential difference between the plates. (discharge) Electric field between the plates is zero. (ii) Only weight and tension in the spring acts on the sphere as electric force is now zero. The net force on the sphere is the difference between the tension and its weight. At equilibrium position, the tension is equal to the weight of the sphere and thus net force is zero. When the sphere is not at its equilibrium position, the net force acting on the sphere is the difference in the tension in the spring from that at the equilibrium position, which always acts towards the equilibrium position. Hence net force, –kx = ma where x is the displacement from the equilibrium position. This gives us a = (−k/m)x, where acceleration is proportional to the displacement of the sphere from the equilibrium position and is always directed opposite to its displacement. This is the defining equation for simple harmonic motion. Sphere oscillates in simple harmonic motion.