Bearing potential of Ball= 5760 V
Explanation:
the electric potential is given by
Q= electrostatic constant= 9 x 10⁹ Nm²/C
Q= charge= n e
n= number of electrons= 2 x 10⁹
e= charge of electron=-1.6 x 10⁻¹⁹ C
so Q=2 x 10⁹ (-1.6 x 10⁻¹⁹ )
Q= -3. 2 x 10⁻¹⁰ C
r= radius of sphere=1/2 mm= 0.0005 m
V= 5760 volt
The total electrostatic force on charge A is
Explanation:
The magnitude of the electrostatic force between two charges is given by Coulomb's law:
where:
is the Coulomb's constant
are the two charges
r is the separation between the two charges
Here we have three positively charged particles A,B and C, located at the following positions:
The magnitudes of the three charges are:
The force exerted by B on A is to the left (because the force between two positive charges is repulsive), and the force exerted by C on A is also to the left (also repulsive). Therefore, the net force on A is just the sum of the two forces exerted by charges B and C:
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