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A rod 12.0 cm long is uniformly charged and has a total charge of -20.0 µc. determine the magnitude and direction of the electric field along the axis of the rod at a point 32.0 cm from its center.

Respuesta :

LearnGrow
Q is the charge and it is Q = -20.0 µC.
Let D denotes the the distance between the center of the rod and the point.Then,
D=0.32 - 0.12 = 0.2 m
L = 0.12 m - the length of the rod
The magnitude and direction of the electric field along the axis of the rod at a point 32.0 cm from its center can be obtained with the formula: 
E = K·Q/r²
E = kQ/D(D+L), where k is a constant with a value of 8.99 x 109 N m2/C2.
So,
E=(8.99 x 109 N m2/C2.* (-20.0 µC))/(0.2 m*0.32m)


barnuts

The formula for the electric field generated by a charged rod is:

E = k Q / [D (D + L)]

Where,

E = electric field

k = Coulomb’s constant = 9 * 10^9 N m^2 / c^2

Q = total charge = -20.0 µc = -20.0 * 10^-6 c

L = length of rod = 12.0 cm = 0.12 m

D = distance from the center = 32.0 – 12.0 cm = 0.20 m

 

Substituting the known values to the formula:

E = (9 * 10^9 N m^2 / c^2) * (-20.0 * 10^-6 c) / [0.20 m (0.20 m + 0.12 m)]

E = -2,812,500 N

Since E is negative therefore E is pointing towards the center of the rod.

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