Answer:
second force = 32.784
Magnitude =
θ = -90°
Explanation:
a)
Fnet = ma
F1 + F2 = ma
20N + F2 = 2(12 × cos30° + 12 ×sin30°)
F2 = 2 × 12 ( sin 30° + cos 30°)
= 24 × ( 1 + √3 )÷ 2
=12 (1 +√3 )
= 32.784
b)
=
=
=
c)
θ = 30° + 180°
θ = 210°
210° - 300°
θ = -90°
Answer:
Net electric field,
Explanation:
Given that,
Charge 1,
Charge 2,
distance, d = 3.2 cm = 0.032 m
Electric field due to charge 1 is given by :
Electric field due to charge 2 is given by :
The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :
So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.
Answer:
U = 1794.005 × 10⁶ J
Explanation:
Data provided;
Capacitance of the original capacitor, C = 1.27 F
Potential difference applied to the original capacitor, V = 59.9 kV
= 59.9 × 10³ V
Now,
The Potential energy (U) for the capacitor is calculated as:
Potential energy of the original capacitor, U = × C × V²
on substituting the respective values, we get
U = × 1.27 × ( 59.9 × 10³ )²
or
U = 1794.005 × 10⁶ J