Resultant force is therefore 0.335 N (F1) (Right→) − 0.182 N (F2) (Left ←) = 0.153 N away from the +3 μC charge (Right→).
5. If two equally charged pith balls are 30 mm apart in air, they cause a repulsive force of 0.04 mN. Calculate the electrostatic charge on each ball.
Solution F = 1
______ 4π ε0
_______ d2
Q1 Q2 4 × 10−5 = 4 × 10−5 = ( (
√ 4 × 10−18 = Q ± 2 nC = Q
___________ You may now complete Exercise 20B (page 238).
CHAPTER 20 EXERCISE 20A HPermittivity of free space = 8.9 × 10–12 F m–1
1. If the relative permittivity of a material is 16.5, calculate the actual permittivity of the material.
2. The permittivity of water is 7.05 × 10−10 F m−1. Calculate the relative permittivity of water.
__________________ 4π (8.9 × 10−12)
1
__________________ 1.118 × 10−10
1
4 × 10−5 = 1 × 1013 Q2 4 × 10−18 = Q2
) ( ) (
_______________ (30 × 10−3)2
(Q1) (Q2)
___________ 9 × 10−4
Q2 ) ) ) Solution E1 +7 µC 1 m halfway 2 m E2 FIG. 20.21 30 mm FIG. 20.20
______ 4π ε0
← E1 (exerted from +15 μC): 1
________ d2 ⇒ (8.9413 × 109)
(Q)(1)
______ 4π ε0
→ E2 (exerted from +7 μC): 1
________ d2 ⇒ (8.9413 × 109)
(Q)(1)
Result at halfway point (←)
(→)
1.3412 × 105 – 6.2589 × 104 = 7.153 N C–1 to the left
1 m +15 µC
6. Calculate the magnitude and direction of the electric field strength halfway between a +7 μC and a +15 μC charge, 2 m apart.
( (
_________________ (1)2
(15 × 10–6)(1) = 1.3412 × 105 N C–1
)
_______________ (1)2
(7 × 10–6)(1) = 6.2589 × 104 N C–1 )
3. If two point charges of −3 μC and +2 μC are placed 25 cm apart in air, what is the magnitude and direction of the force on the +2 μC charge?
4. A space contains two point charges of −6 μC and +6 μC, 2 metres apart in a vacuum. Calculate the direction and magnitude of the force on the +6 μC charge.
