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Electric Field and Gauss Law
416335 Electric Field and Gauss Law 4) A charged ball with charge l0 Coulombs is placed at (0,0,0). A charged ball with
charge -5 Coulombs is placed at (2,0,0).
a) Where do you place a +2 Coulomb ball so it will feel no net force?
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Charged wall and hanging charged plastic ball
Draw and label the forces acting on the ball (diagram included in attachment).
b.) Calculate the magnitude of the electric field at the ball's location due to the charged wall, and state its direction relative to the coordinate axes shown.
c.)
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A small plastic ball of mass is suspended between capacitor plates: What is the magnitude of the charge on each plate?
Therefore there is a constant electric field between the plates.
Since the ball is charged and placed in an electric field, the field applies a force F.
This force, together with gravity is balanced by the tension in the string.
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Charged pendulum in electric field
48627 Charged pendulum in electric field A 1.00 g cork ball with charge 2.00 uC is suspended vertically on a 5.00 m long, light string in the presence of a uniform, downward-directed electric field of magnitude E = 1.00 x 10^5 N/C.
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Electric field due to a uniformly charged Sphere
Find the magnitude of the electric field at a point 1.5×10^−2 m from the center of the sphere.
We have a spherically charged sphere. We are asked to find the electric fields inside, on and out side of the sphere.
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Charged concentric spheres
Electric field on the Gaussian surface due to the charge on the solid sphere will be uniform and given by the equation :
Ea = kQa/r2 = 9x109x4.5x10-6/r2 = 40.5x103/r2
As Ea = - dVa/dr, Va = - ∫Eadr = - 40.5x103 ∫1/r2 dr = + (40.5x103)1/r
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potential at the center of the charged solid sphere
202157 Find potential at center of charged, solid sphere The electric field at the surface of a charged, solid, copper sphere with radius 0.250 m is 3400 N/C , directed toward the center of the sphere.
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3 Q Electric Field: Point Charges, Ring, Sphere, Gauss,
Reading
Answer: The electric field due to a ring charged uniformly with charge Q at a point distance a from the center on the axis of the ring is given by
Now if we want field at the center of the ring then a = 0
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Working with an electroscope
Thus the horizontal forces keeping the balls apart and hence the leaves are the Coulomb forces due to each charged ball.