Question
A ball of mass $$1\,g$$ carrying a charge $${10^{ - 8}}C$$ moves from a point $$A$$ at potential $$600\,V$$ to a point $$B$$ at zero potential. The change in its K.E. is
A.
$$ - 6 \times {10^{ - 6}}\,erg$$
B.
$$ - 6 \times {10^{ - 6}}\,J$$
C.
$$6 \times {10^{ - 6}}\,J$$
D.
$$6 \times {10^{ - 6}}\,erg$$
Answer :
$$6 \times {10^{ - 6}}\,J$$
Solution :
As work is done by the field, $$K.E.$$ of the body increases by
$$\eqalign{
& K.E. = W = q\left( {{V_A} - {V_B}} \right) \cr
& = {10^{ - 8}}\left( {600 - 0} \right) \cr
& = 6 \times {10^{ - 6}}\,J \cr} $$