Current is produced when charges are accelerated by an electric field to move to a position of lower
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Answer:
The focal length of the appropriate corrective lens is 35.71 cm.
The power of the appropriate corrective lens is 0.028 D.
Explanation:
The expression for the lens formula is as follows;
Here, f is the focal length, u is the object distance and v is the image distance.
It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.
Put v= -71.4 cm and u= 24.0 cm in the above expression.
f= 35.71 cm
Therefore, the focal length of the corrective lens is 35.71 cm.
The expression for the power of the lens is as follows;
Here, p is the power of the lens.
Put f= 35.71 cm.
p=0.028 D
Therefore, the power of the corrective lens is 0.028 D.
Answer:
Explanation:
The electric field produced by a single point charge is given by:
where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
In this problem, we have
E = 1.0 N/C (magnitude of the electric field)
r = 1.0 m (distance from the charge)
Solving the equation for q, we find the charge: