INDUCTION MOTOR:-
Speed:-Less speed range than PMAC motors • Speed range is a function of the drive being used — to 1,000:1 with an encoder, 120:1 under field-oriented control
Reliability:-Waste heat is capable of degrading insulation essential to motor operation • Years of service common with proper operation
Power density:-Induction produced by squirrel cage rotor inherently limits power density
Accuracy:-Flux vector and field-oriented control allows for some of accuracy of servos
Cost:-Relatively modest initial cost; higher operating costs
PERMANENT MAGNET MORTOR:-
speed:-VFD-driven PMAC motors can be used in nearly all induction-motor and some servo applications • Typical servomotor application speed — to 10,000 rpm — is out of PMAC motor range
Reliability:-Lower operating temperatures reduces wear and tear, maintenance • Extends bearing and insulation life • Robust construction for years of trouble-free operation in harsh environments.
power density:-Rare-earth permanent magnets produce more flux (and resultant torque) for their physical size than induction types.
Accuracy:-Without feedback, can be difficult to locate and position to the pinpoint accuracy of servomotors
<span>Cost:-Exhibit higher efficiency, so their energy use is smaller and full return on their initial purchase cost is realized more quickly</span>
The electric potential between the two charges is 91.68 V.
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Electric potential between the two charges</h3>
The electric potential between the two charges is calculated as follows;
V = Ed
where;
- V is electric potential
- E is electric field
- d is the distance of the charge
Substitute the given parameters and solve for electric potential,
V = 573 N/c x 0.16 m
V = 91.68 V
Thus, the electric potential between the two charges is 91.68 V.
Learn more about electric potential here: brainly.com/question/26978411
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Answer:
Electric field, E = 45.19 N/C
Explanation:
It is given that,
Surface charge density of first surface, 
Surface charge density of second surface, 
The electric field at a point between the two surfaces is given by :
E = 45.19 N/C
So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.
