The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
<h3>Electric field on the master charge</h3>
E = kq/r²
where;
- q is magnitude of master charge
- r is distance of separation
- k is Coulomb's constant
E = (9 x 10⁹ x 0.63)/(0.75²)
E = 1.008 x 10¹⁰ N/C
<h3>Force on the test charge</h3>
F = Eq
where;
- E is electric field
- q is the test charge
F = (1.008 x 10¹⁰) x (0.5)
F = 5.04 x 10⁹ N
Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
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Answer:
F = 4.147 × 10^23
v = 1.31 × 10^4
Explanation:
Given the following :
mass of Jupiter (m1) = 1.9 × 10^27
Mass of sun (m2) = 1.99 × 10^30
Distance between sun and jupiter (r) = 7.8 × 10^11m
Gravitational force (F) :
(Gm1m2) / r^2
Where ; G = 6.673×10^-11 ( Gravitational constant)
F = [(6.673×10^-11) × (1.9 × 10^27) × (1.99 × 10^30)] / (7.8 × 10^11)^2
F = [25.231 × 10^(-11+27+30)] / (60.84 × 10^22)
F = (25.231 × 10^46) / (60.84 × 10^22)
F = 3.235 × 10^(46 - 22)
F = 0.4147 × 10^24
F = 4.147 × 10^23
Speed of Jupiter (v) :
v = √(Fr) / m1
v = √[(4.147 × 10^23) × (7.8 × 10^11) / (1.9 × 10^27)
v = √32.3466 × 10^(23+11) / 1.9 × 10^27
v = √32.3466× 10^34 / 1.9 × 10^27
v = √17. 023 × 10^34-27
v = √17.023 × 10^7
v = 13047.221
v = 1.31 × 10^4
The physical law that explains that is the law of conservation of energy which states that he energy of an isolated sistem remains constant
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