DHS 15 Electric Fields (Tutorial Solutions)

Dunman High School (Senior High Physics Department) Year 6 Physics H2 Tutorial Topic 13: Electric Fields Tutorial Solutions 2025 9749 Physics (2025) Topic 13: Electric Fields For Internal Use Only Page 1 of 15 Concept of an electric field 1 (a) Q1 = +2.0 C Q2 = 1.0 C (b) Q1 = +2.0 C Q2 = 3.0 C (c) Q1 = +2.0 C Q2 = +1.0 C

Dunman High School (Senior High Physics Department) Year 6 Physics H2 Tutorial Topic 13: Electric Fields Tutorial Solutions 2025 9749 Physics (2025) Topic 13: Electric Fields For Internal Use Only Page 2 of 15 (d) Q1 = +2.0 C Q2 = +3.0 C 2 Electric Gravitational constant 0 1 4 G field strength FE q Fg m field strength of a point charge Q or point mass M 2 0 1 4 QE r 2 Mg Gr force between point charges (masses) 1 2 2 0 1 4 Q QF r (Coulomb’s law) 1 2 2 m mF G r (Newton’s law of gravitation) potential WV q W m potential of a point charge Q or point mass M 0 1 4 QV r GM r field strength at a point is numerically equal to the potential gradient at that point dVE dr dg dr  Point charges 3 Based on 2 0 1 4 QE r and 0 1 4 QV r .

Dunman High School (Senior High Physics Department) Year 6 Physics H2 Tutorial Topic 13: Electric Fields Tutorial Solutions 2025 9749 Physics (2025) Topic 13: Electric Fields For Internal Use Only Page 3 of 15 4 (a) 2 0 21 4 r QQF  (b) r QVA 0 2 4 (c) 1 2 AB 04 Q QU r (d) r QVB 0 1 4 (e) (f) Q1 and Q2 are point charges; The separation between the charges is much larger than the dimensions of the charges; The charges are uniformly distributed on the object surfaces. 5 (a)       o o 2 2 0 0 o 2 0 vector sum of fields due to +1C & 1C, to the right vector sum of fields due to +2C & 2C, to the right 1 1 1 22 cos 45 2 cos 454 42 0.5 2 0.5 1 13 cos 454 0.5 E E                           = 7.64  1010 N C−1 towards the right. (b) = 0 J C−1 6 Electric field strength at the fourth corner due to the two +Q charges are E1 and E2, with magnitude 24 o Q r . E4 is the resultant of E1 and E2, hence E4 = o Q r 2 2 4 . Electric field strength due to the Q charge is E3 = 24 ( 2)o Q r = 28 o Q r . Since E3 is less than E4, the direction is B. The magnitude of E4 and E3 is o Q r 2 2 4 − 28 o Q r = 9 1 2 2 2 1 8.22 10 N C8 o Q r    r QQQQV rr 0 21 20 2 20 1 244   5.0 )2)(109( 5.0 )2)(109( 5.0 )1)(109( 5.0 )1)(109( 9999  E2 E4 E3 E1

Dunman High School (Senior High Physics Department) Year 6 Physics H2 Tutorial Topic 13: Electric Fields Tutorial Solutions 2025 9749 Physics (2025) Topic 13: Electric Fields For Internal Use Only Page 4 of 15 Uniform electric fields 7 (Answer: D) Electric field strength between parallel plates is constant throughout. 8 (2007 P1 Q24) (Answer: B) The weight acts downwards and has to be considered. The sphere is suspended stationary, therefore it is in equilibrium. To counter the weight, we need an upward force. Find the combination of charges and polarity that achieves this. X and Y has to be oppositely-charged to attract, and Y and Z should be like-charged to repel. F W

Dunman High School (Senior High Physics Department) Year 6 Physics H2 Tutorial Topic 13: Electric Fields Tutorial Solutions 2025 9749 Physics (2025) Topic 13: Electric Fields For Internal Use Only Page 5 of 15 9 (a) Apply = VE d 2 1 10 V (5.0 10 ) m 200 V m (vertically downwards) E      (b) Magnitude of electric force, F = qE = (2.0 × 10−6)(200) = 4.0 × 10−4 N Direction of electric force: vertically upwards (c) Force on negative charge acts towards the plate with a higher electric potential. Notice the electric force exerted on a negative charge acts opposite to the direction of electric field between the plates. 10 Magnitude of force = |gradient of graph| = 161.5 10 0 0 0.020    = 7.5 × 10−15 N 11 F E high potential plate low potential plate

Dunman High School (Senior High Physics Department) Year 6 Physics H2 Tutorial Topic 13: Electric Fields Tutorial Solutions 2025 9749 Physics (2025) Topic 13: Electric Fields For Internal Use Only Page 6 of 15 F = qE = q V d  = (5.0  10-6) 2 100 40 2.0 10      = 1.5 × 10−2 N (in direction of the electric field) 12 (a) Horizontally, the electron moves with uniform velocity ux; using sx = uxt, time interval that the electron is inside the electric field is t = x x s u = . 7 0 10 2 10 = 5.0 × 10−9 s (b) Vertically, the electron moves with uniform acceleration a: ma = qE = e| |V d   a = eV md =     . . . 19 31 2 1 60 10 250 9 11 10 5 0 10       = 8.78 × 1014 m s−2 (vertically upwards) (c) Using sy = uyt + 1 2ayt2, sy =1 2ayt2 = eV md2     2 x x s u =     . . .. . 19 2 731 2 1 60 10 250 0 10 2 0 102 9 11 10 5 0 10           = 0.011 m (d) The direction of the electric field strength (high V to low V): downwards. The direction of the electric force acting on an electron (ve): upwards. (e) (i) F/N 8.0  10−16 0 5 x/cm (ii) V/V 250 0 5 x/cm (iii) E/ V m−1 0 5 x/cm −5000

Dunman High School (Senior High Physics Department) Year 6 Physics H2 Tutorial Topic 13: Electric Fields Tutorial Solutions 2025 9749 Physics (2025) Topic 13: Electric Fields For Internal Use Only Page 7 of 15 13 (a) 7500 × 1.60 × 10–19 = 9.11 × 10–31 a a = 1.317 × 1015 m s–1, downwards The electric field lines should be straight and equally spaced parallel lines directed upwards. The path of the beam is a parabolic curve. (b) Let the speed of electrons emerging from B be u and their time of travel between A and B be T. sx = ux t: 0.025 = (u cos 15˚) T o 0.025 cos15T u ––––––––– (1) When the beam is at the highest point, vy = uy + ayt: 0 = u sin 15˚– (1.317 × 1015) 2 T    u sin 15˚ = (1.317 × 1015) 2 T    ––––––––– (2) Substitute (1) into (2): u sin 15˚ = (1.317 × 1015) o 0.025 2 cos 15u     u = 8.11 × 106 m s –1 Electric potential energy 14  system 1 2 2 3 1 3 0 1 4 0.798 J EPE q q q q q qr        system 0 1in general, total 4 i j i j ij q qEPE r      -145 7500 N C0.006E  Newton's 2nd law: F ma qE ma  

Dunman High School (Senior High Physics Department) Year 6 Physics H2 Tutorial Topic 13: Electric Fields Tutorial Solutions 2025 9749 Physics (2025) Topic 13: Electric Fields For Internal Use Only Page 8 of 15 15 Conservation of energy:          2 2 2 0 4 2 10 0 4 25 10 5 0 i 0 .e. incre 0 1 0 02 1 02 1 1 1 02 4 1 1.0 10 1 12.0 1.5 10 02 4 0.1 1. 1 1.0 102.0 1.5 10 9 2 4 10 10 electric f i f i K U mv q V mv q V V Qmv q r r v v                                                                1 ase in KE = decrease i 1 n EPE 1 m sv   16 Along