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Damm [24]
3 years ago
15

Two plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.50

nC . Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 900-mm stick stuck through the holes so that it runs from the center of one ball to the center of the other.
Required:
What is the magnitude of the dipole moment of the arrangement?
Physics
2 answers:
Sergio039 [100]3 years ago
7 0

Answer:

The dipole moment is 4.5 x 10^-10 Cm.

Explanation:

Charge on each ball, q = 0.5 nC

Length, L = 900 mm = 0.9 m

The dipole moment is defined as the product of either charge and the distance between them.

It is a vector quantity and the direction is from negative charge to the positive charge.

The dipole moment is

p = q L\\\\p = 0.5 \times 10^{-9}\times 0.9\\\\p = 4.5\times 10^{-10} Cm

galben [10]3 years ago
5 0

Answer:

The right solution is "4.5\times 10^{-10} \ Cm".

Explanation:

Given that,

q = 0.50 nC

d = 900 mm

As we know,

⇒ P=qd

By putting the values, we get

⇒     =0.50\times 900

⇒     =(0.50\times 10^{-9})\times 0.9

⇒     =4.5\times 10^{-10} \ Cm  

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