the total electric potential at location P, which is at the center of the rectangle is 0V.
The charges placed at the corner of the rectangle are same in magnitude but different in charge. hence the total electric potential will be same in magnitude but different in charge and will be cancelled. Similarly, all the total electric potential will be cancelled and resultant will be zero.
<h3>
What is total electric potential?</h3>
- The amount of labor required to convey a unit of electric charge from a reference point to a given place in an electric field is known as the electric potential (also known as the electric field potential, potential drop, or the electrostatic potential).
- More specifically, it is the energy per unit charge for a test charge that is negligibly disruptive to the field under discussion. In order to prevent the test charge from gaining kinetic energy or radiating, the travel across the field is also meant to occur with very little acceleration.
- The electric potential at the reference location is, by definition, zero units. Any point may be used as the reference point, but typically it is earth or a point at infinity.
To learn more about total electric potential with the given link
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Answer:
0.0061 J
Explanation:
Parameters given:
Number of turns, N = 111
Radius of turn, r = 2.11 cm = 0.0211 m
Resistance, R = 14.1 ohms
Time taken, t = 0.125 s
Initial magnetic field, Bin = 0.669 T
Final magnetic field, Bfin = 0 T
The energy dissipated in the resistor is given as:
E = P * t
Where P = Power dissipated in the resistor
Power, P, is given as:
P = V² / R
Hence, energy will be:
E = (V² * t) / R
To find the induced voltage (EMF), V:
EMF = [-(Bfin - Bin) * N * A] / t
A is Area of coil
EMF = [-(0 - 0.669) * 111 * pi * 0.0211²] / 0.125
EMF = 0.83 V
Hence, the energy dissipated will be:
E = (0.83² * 0.125) / 14.1
E = 0.0061 J
Answer:
m = 0.25
Explanation:
Given that,
Object distance, u = -15cm
Height of the object, h = 48
Focal length, f = cm
We need to find the magnification of the image.
Let v is the image distance. Using mirror's equation.
Magnification,
Hence, the magnification of the image is 0.25.
Answer:
b . Move the conductor near a magnet
