Answer:
CB = 4.45 x 10⁻⁹ F = 4.45 nF
Explanation:
The capacitance of a parallel plate capacitor is given by the following formula:
C = ε₀A/d
where,
C = Capacitance
ε₀ = Permeability of free space
A = Area of plates
d = Distance between plates
FOR CAPACITOR A:
C = CA = 17.8 nF = 17.8 x 10⁻⁹ F
A = A₁
d = d₁
Therefore,
CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1
FOR CAPACITOR B:
C = CB = ?
A = A₁/2
d = 2 d₁
Therefore,
CB = ε₀(A₁/2)/2d₁
CB = (1/4)(ε₀A₁/d₁)
using equation 1:
CB = (1/4)(17.8 X 10⁻⁹ F)
<u>CB = 4.45 x 10⁻⁹ F = 4.45 nF</u>
Answer:
Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from
The answer to the question is
The speed
of the electron when it is 10.0 cm from charge Q₁
= 7.53×10⁶ m/s
Explanation:
To solve the question we have
Q₁ = 3.45 nC = 3.45 × 10⁻⁹C
Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C
2·d = 50.0 cm
a = 10.0 cm
q = -1.6×10⁻¹⁹C
Also initial kinetic energy = 0 and
Initial electric potential energy = 
Final kinetic energy due to motion = 0.5·m·v²
Final electric potential energy = 
From the energy conservation principle we have

Solving for v gives

where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg
gives v =7528188.32769 m/s or 7.53×10⁶ m/s
= 7.53×10⁶ m/s
Answer:
Explanation:
The change in force = 9.81(1.95 - 0.300) = 16.2 N
The change in length is 0.750 - 0.200 = 0.550 m
K = ΔF/Δx = 16.2/0.550 = 29.4 N/m
Answer:
T = 1.73 kg
Explanation:
Let's use Newton's second law for this balance problem
We draw a coordinate system with the x axis parallel to the plane and the Y axis perpendicular
The only force we have to lay down is the weight (W)
Wx = W sin θ
Wy = W cos θ
Wx = 2.0 ain 60
Wx = 1.73 kg
Wy = 2.0 cos 60
Wy = 1.0 kg
Y Axis
N-Wy = 0
X axis
T- Wx = 0
T = Wx
T = 1.73 kg