Two point charges qA = 3 μC and qB = –3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 x 10–9 C is placed at this point, what is the force experienced by the test charge?
and
(a)
Let us assume that a unit positive test charge is placed at 0. qA will repel this test charge while qB will attract. Hence, both are directed towards
(b) As a negative test charge of q0 = – 1.5 x 10–6 C is placed at 0. qA will attract it while qB will repel. Therefore, the net force
Check that the ratio is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?
and
Now,
It also establishes that the electrostatic force is about 1039 times stronger than the gravitational force.
Law of conservation of charge states that total charge on an isolated system of objects always remain conserved.
When a glass rod is rubbed with silk cloth, glass rod becomes positively charged while silk cloth becomes negatively charged and, the amount of positive charge on the glass rod is apparently found to be exactly the same as the negative charge on silk cloth. Since, the measure of charge is same on both, equal amount of charge with opposite nature will cancel out each other. Hence, the total sum of charge on two bodies is zero. Thus, the system of glass rod and silk cloth, which was neutral before rubbing, still possesses no net charge after rubbing. And law of conservation of charge is justified.
As a consequence of conservation of charge, when two charged conductors of same size and same material carrying charges Q1 and Q2 respectively are brought in contact and separated, the charge on each conductor will be
This condition, however, does not hold true if the conductors are of different sizes or of different material.
In that case the charges on the conductors will be Q1' and Q2’ respectively, where Q1 + Q2 = Q1‘ + Q2’.