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?

Respuesta :

Answer:

The right solution is "[tex]4.5\times 10^{-10} \ Cm[/tex]".

Explanation:

Given that,

q = 0.50 nC

d = 900 mm

As we know,

⇒ [tex]P=qd[/tex]

By putting the values, we get

⇒     [tex]=0.50\times 900[/tex]

⇒     [tex]=(0.50\times 10^{-9})\times 0.9[/tex]

⇒     [tex]=4.5\times 10^{-10} \ Cm[/tex]  

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

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