The electric potential V(z) on the z-axis is : V =
The magnitude of the electric field on the z axis is : E = kб 2( 1 - [z / √(z² + a² ) ] )
<u>Given data :</u>
V(z) =2kQ / a²(v(a² + z²) ) -z
<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>
Considering a disk with radius R
Charge = dq
Also the distance from the edge to the point on the z-axis = √ [R² + z²].
The surface charge density of the disk ( б ) = dq / dA
Small element charge dq = б( 2πR ) dr
dV ----- ( 1 )
Integrating equation ( 1 ) over for full radius of a
∫dv =
V =
=
Therefore the electric potential V(z) =
Also
The magnitude of the electric field on the z axis is : E = kб 2( 1 - [z / √(z² + a² ) ] )
Hence we can conclude that the answers to your question are as listed above.
Learn more about electric potential : brainly.com/question/25923373
Answer:
100% efficiency
Explanation:
the machine has _100%_ efficiency.
(fill in the blank)
Answer:
9.25 x 10^-4 Nm
Explanation:
number of turns, N = 8
major axis = 40 cm
semi major axis, a = 20 cm = 0.2 m
minor axis = 30 cm
semi minor axis, b = 15 cm = 0.15 m
current, i = 6.2 A
Magnetic field, B = 1.98 x 10^-4 T
Angle between the normal and the magnetic field is 90°.
Torque is given by
τ = N i A B SinФ
Where, A be the area of the coil.
Area of ellipse, A = π ab = 3.14 x 0.20 x 0.15 = 0.0942 m²
τ = 8 x 6.20 x 0.0942 x 1.98 x 10^-4 x Sin 90°
τ = 9.25 x 10^-4 Nm
thus, the torque is 9.25 x 10^-4 Nm.
92 grams of Y would combine with 46 grams of X
Answer: The correct answer is option C.
Explanation:
Weight = Mass × Acceleration
Let the mass of the space probe be m
Acceleration due to gravity on the earth = g
Weight of the space probe on earth = W
Acceleration due to gravity on the Jupiter = g' = 2.5g
Weight of the space probe on earth = W'
The weight of the space probe on the Jupiter will be 2.5 times the weight of the space probe on earth.
Hence, the correct answer is option C.