The solution for this problem is:
For 1st minimum, let m be equal to 1.
d = slit width
D = screen distance.
Θ = arcsin (m * lambda/ (d))
= 0.13934 rad, 7.9836 deg
y = D*tan (Θ)
y = 6.50 * tan (7.9836)
= 0.91161 m is the distance from the central maximum to the first-order minimum
Answer:
1. 8437500 N
2. The force between the two charges is attractive.
Explanation:
1. Determination of the force between the two charges.
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
Distance apart (r) = 80 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 2 × 3 / 80²
F = 5.4×10¹⁰ / 6400
F = 8437500 N
Thus, the force of attraction between the two charges is 8437500 N
2. From the question given, the charges are:
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
We understood that like charges repels while unlike charges attract. Since the two charges (i.e –2 C and 3 C) has opposite signs, it means they will attract each other.
Thus the force between them is attractive.
When the object is big enough to contract itself into a ball.
