Answer:
Explanation:
Given:
- Three identical charges q.
- Two charges on x - axis separated by distance a about origin
- One on y-axis
- All three charges are vertices
Find:
- Find an expression for the electric field at points on the y-axis above the uppermost charge.
- Show that the working reduces to point charge when y >> a.
Solution
- Take a variable distance y above the top most charge.
- Then compute the distance from charges on the axis to the variable distance y:
- Then compute the angle that Force makes with the y axis:
cos(Q) = sqrt(3)*a / 2*r
- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:
F_1,2 = 2*F_x*cos(Q)
- The total net force would be:
F_net = F_1,2 + kq / y^2
- Hence,
- Now for the limit y >>a:
- Insert limit i.e a/y = 0
Hence the Electric Field is off a point charge of magnitude 3q.
Answer:
100 N
Explanation:
Given that,
Two forces whose resultant is 100newton are perpendicular to each other.
If one of them makes an angle of 60newton with the resultant.
and
The magnitude of force,
or
F = 100 N
So, the magnitude of force is 100 N.
What is it asking? It is sort of blurry
The right answer for the question that is being asked and shown above is that:
B.
Property Electromagnet Magnet
Permanent magnet X
Temporary magnet X X
Magnetic field created by aligning domains X
<span>Moving electric charge produces magnetic field X</span>