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) T