A hollow sphere of negligible mass and radius R is completely filled with a liquid so that its density is rho. You now enlarge t
1 answer:
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The relationship between the frequency (f) of an electromagnetic wave, the speed of the wave (which is the speed of light, c) and its wavelength 
is given by
Since the wavelength of the radiation in the problem is
, by substituting numbers in the formula we can get the frequency:
Since the wavelength of the radiation in the problem is
Answer:
a) 80 V
b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge
Explanation:
<u>Given :</u>
We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( = 0) to the point B at distance l = 0.500m where its kinetic energy is
(
= 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = +
30.0 v.
<u>Required :</u>
<em>(a) We are asked to find the electric potential VB </em>
<em>(b) We want to determine the magnitude and the direction of the electric field E. </em>
<u> Solution </u>
(a) We are given the values for VA, and q so we want to find a relationship between these three parameters and VB to get the value of VB.
As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:
.........................................(1)
Where = 0 and the potential energy U of the charge is given by U = q V
where V is the electric potential. So, equation (1) will be in the form :
0+qVA= +qVB (Divide by q)
VA= /q + VB (solve for VB)
VB=VA- /q .......................................(2)
We get the relation between VB, VA and , now we can plug our values for VA,
and q into equation (2) to get VB
VB=VA- /q
=30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)
=80 V
(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points
VA-VB =...................................(a)
VA-VB=E (solve for E)
E= VA-VB/..................................(3)
Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E
E= VA-VB/
=-100 N/C
The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge
Answer:
The magnitude of the force on the wire is 2.68 N.
Explanation:
Given that,
Length of the wire, L = 5 m
Magnetic field, B = 0.37 T
Angle between wire and the magnetic field,
Current in the wire, I = 2.9 A
We need to find the magnitude of the force on the wire. The magnetic force in the wire is given by :
So, the magnitude of the force on the wire is 2.68 N. Hence, this is the required solution.