Answer:
2500 N/C
Explanation:
Electric Field: This can be defined as the the force per unit charge in an electric field. The S.I unit is N/C
V = E.r ....................... Equation 1
Where V = Electric potential, E = Electric Field, r = Distance between the two parallel plate,
Make E the subject of the equation,
E = V/r .................... Equation 2
Given: V = 10 V, r = 4.00 mm = (4/1000) m = 0.004 m.
Substitute into equation 2
E = 10/0.004
E = 2500 N/C
Hence the electric field = 2500 N/C
Because the weight of one ball is mg = 147 N, the gravitational force between the two balls is or parts per billion of the weight.
Answer:
Friction force is 0.1375 N
Solution:
As per the question:
Radius of the metal disc, R = 4.0 cm = 0.04 m
Magnetic field, B = 1.25 T
Current, I = 5.5 A
Force on a current carrying conductor in a magnetic field is given by considering it on a differential element:
Integrating the above eqn:
(1)
Now the torque is given by:
(2)
From eqn (1) and (2):
Thus the Frictional force is given by: