Answer:
<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>
Explanation:
The initial charge on 4 mF capacitor = 4 mf x 50 V = 200 mC
The initial Charge on 6 mF capacitor = 6 mf x 30 V =180 mC
Since the negative ends are joined together the total charge on both capacity would be;
q = 
q = 200 - 180
q = 20 mC
In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage
q = (4 x V) + (6 x V)
20 = 10 V
V = 2 V
For the final charge on 6.0 mF;
q = CV
q = 6.0 mF x 2 V
q = 12 mC
Therefore the final charge on the 6.0 mF capacitor would be 12 mC
Answer:
Explanation:
Given that
J(r) = Br
We know that area of small element
dA = 2 π dr
I = J A
dI = J dA
Now by putting the values
dI = B r . 2 π dr
dI= 2π Br² dr
Now by integrating above equation
Given that
B= 2.35 x 10⁵ A/m³
r₁ = 2 mm
r₂ = 2+ 0.0115 mm
r₂ = 2.0115 mm
By putting the values
Answer:
18 W
Explanation:
Applying,
P = V²/R.................. Equation 1
Where P = Power of both glowing bulbs, V = Voltage, R = Combined Resistance of both bulbs
Since: It is a series circuit,
Then,
R = R1+R2............. Equation 2
Where R1= Resistance of the first bulb, R2 = Resistance of the second bulb
Given: R1 = R2 = 8 Ω
Substitute into equation 1
R = 8+8
R = 16 Ω
Also Given: V = 12 V
Substitute into equation 1
P = 12²/8
P = 144/8
P = 18 W
Answer:
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
Explanation:
I cannot find any attached photo, but we can proceed anyways theoretically.
The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point
i.e
But the force F
But the electric field intensity due to a point charge Q at a distance r meters away is given by
<em>From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases</em>
Answer:
Hence, work done= 287.54 J
Explanation:
Given data:
angle of ramp with the ground θ =20°
force applied = 76 N
work done on the crate to slide down 4 m down the ramp
W= F×d cosθ ( only the cos component of the force will slide the crate down)
W= 76×4×cos20= 287.54 J
