Download reasoning and solution

International Indian School, Riyadh, K.S.A.
Mohammad Abdul Mumeed, Senior Secondary Teacher in Physics
Electric Forces & Electric Fields(With Reasoning & Solution)
These questions contain both Concepts & Calculations and are designed to
enhance the subject knowledge of the students. Advised to work alone or in
a small group before checking the Solution i.e. for Horizontal- learning
Numerical Question#1:
Two very small spheres are initially neutral and separated by a distance of
0.50m. Suppose that 3.0x10 13 electrons are removed from one sphere and
placed on the other.
a) What is the magnitude of the electrostatic force that acts on each
sphere?
b) Is the force attractive or repulsive? Why?
REASONING AND SOLUTION
Initially, the two spheres are neutral. Since negative charge is removed from
the sphere which loses electrons, it then carries a net positive charge.
Furthermore, the neutral sphere to which the electrons are added is then
negatively charged. Once the charge is transferred, there exists an
electrostatic force on each of the two spheres, the magnitude of which is
given by Coulomb's law
SOLUTION
Since each electron carries a charge of
the amount of negative charge removed from the first sphere is
Thus, the first sphere carries a charge
while the second sphere carries a charge
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International Indian School, Riyadh, K.S.A.
Mohammad Abdul Mumeed, Senior Secondary Teacher in Physics
The magnitude of the electrostatic force that acts on each sphere is,
therefore,
b. Since the spheres carry charges of opposite sign, the force is
attractive.
Numerical Question#2:
A tiny ball of mass 0.012 kg carries a charge of -18 µC. What electric field
(magnitude and direction) is needed to cause the ball to float above the
ground?
REASONING AND SOLUTION
Two forces act on the charged ball (charge q); they are the downward force
of gravity mg and the electric force F due to the presence of the charge q in
the electric field E. In order for the ball to float, these two forces must be
equal in magnitude and opposite in direction, so that the net force on the ball
is zero (Newton's second law). Therefore, F must point upward, which we
will take as the positive direction. According to Equation
F = qE. Since the charge q is negative, the electric field E must point
downward, as the product qE in the expression F = qE must be positive,
since the force F points upward. The magnitudes of the two forces must be
equal, so that
. This expression can be solved for E.
SOLUTION The magnitude of the electric field E is
As discussed in the reasoning, this electric field points downward.
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