Question

Two plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.60 nC . Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 800-mm stick stuck through the holes so that it runs from the center of one ball to the center of the other.What is the magnitude of the dipole moment of the arrangement?

Answer 1

The electric dipole moment for a pair of opposite charges of magnitude q is defined as the magnitude of the charge times the distance between them and the defined direction is toward the positive charge, mathematically can be written as,

$P= qd$

Here,

q= Chage

d= Distance

Our values are given as,

$q = 0.60nC$

$d = 800mm$

Replacing,

$P= (0.60 nC)*( 800mm)$

$P= (0.60*10^{-9}C)*(0.8m)$

$P= 4.8*10^{-10} C*m$

Therefore the magnitude of the dipole moment of the arrangement is $4.8*10^{-10} C*m$

Answer 2

The magnitude of the dipole moment of the given arrangment of the balls is $4.8 \times 10^{-9} \rm Cm$

The electric dipole moment:

Its magnitude can be defined as the product of the charge and distance between them.

$p = qd$  

Where,

$q$- Chage  =  0.60 nC = $0.6 \times 10^{-9} \rm C$

$d$ -  Distance = 800-mm  

Put the values in the formula,

$p = 0.6 \times 10^{-9} \times 0.8 m \\\\p = 4.8 \times 10^{-9} \rm Cm$

Therefore, the magnitude of the dipole moment of the given arrangement of the balls is $4.8 \times 10^{-9} \rm Cm$

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