Solution:
Solution
A charge $$=+Q$$
mass $$=m$$
magnitude $$=+2Q$$
mass $$=2m$$
Velocity $$={ V }_{ 0 }$$
Using coulomb's law, charge $$Q$$ and thus the force
$$F=K\dfrac { { Q }^{ 2 } }{ { d }^{ 2 } } =0$$ and it will become $$O$$ because it is inner surface.
Now, force applied by floor on the fixed charge in horizontal direction
$$F=\dfrac { K\left( 2Q \right) \left( Q \right) }{ { d }^{ 2 } } =\dfrac { K\times { 2Q }^{ 2 } }{ { d }^{ 2 } } =\dfrac { 2K{ Q }^{ 2 } }{ { d }^{ 2 } } $$
