The arrows around the positive point charge illustrates the direction of the electric force/field.
<h3>What is Electric field?</h3>
This is described as an area in which force is exerted om other charged particles or objects.
Arrows could be around the positive or negative point charge and helps to depict the direction of the electric force.
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Answer:
the sheets approach while the object is near
Explanation:
An electroscope is an apparatus that has two metal sheets attached, when these sheets have a charge and distribute evenly between them and the sheets repel.
When I approach an object charged with a counter (negative) charge, part of the charge of the electroscope moves near the charged external object, to neutralize the electric field, so as the charge on one of the plates decreases the electroscope has approached , as the objects are not touched the system remains in this configuration while the object is close. When the object is released, the electric field it creates disappears, so the positive charges repel inside the electroscope and the sheets repel to the initial position.
In short, the sheets approach while the object is near
Answer:
23.5
Explanation:
Dunno how 2 explain but this is correct 4 sureeeeee.
Answer:
Explanation:
Length = 1.00 m
If the length is 1.0, the vertical distance pivot to bob is cos 35 = 0.819
At the lowest point, vertical distance is 1.0, so the change is the difference, 0.181 meter
The potential energy of that height is converted to kinetic energy of motion, which determines the speed.
PE = KE
mgh = ½mV²
V = √(2gh) = 1.88 m/s
Answer:
E_aprox = 1.003 E_real
Explanation:
In this exercise we are given the expression for the electric field of a dipole in the axis direction of the dipole
E_real = k 2q d / √(z² + d²)³
I think your equation has some errors.
In this case they indicate that d is the separation of the charges of the dipole
in the case of z »d this equations approximates
E_aprox = k 2q d/ z³
calculate the value for the two cases
E_real = k2q d / √[ ((23d)² + d²)³]
E_real = k2q d / d³ 12201
E_real = k2q 1/12201 d²
E_aprox = k2q d / (23.00d)³
E_aprox = k2q 1/12167 d²
the error between these quantities is
E_aprox / E_real = 12201 d² / 12167 d²
E_aprox / E_real = 1.003
E_aprox = 1.003 E_real