Question
In moving from $$A$$ to $$B$$ along an electric field line, the work done by the electric field on an electron is $$6.4 \times {10^{ - 19}}J.$$ If $${\phi _1}$$ and $${\phi _2}$$ are equipotential surfaces, then the potential difference $${V_C} - {V_A}$$ is
In moving from $$A$$ to $$B$$ along an electric field line, the work done by the electric field on an electron is $$6.4 \times {10^{ - 19}}J.$$ If $${\phi _1}$$ and $${\phi _2}$$ are equipotential surfaces, then the potential difference $${V_C} - {V_A}$$ is
A.
$$-4\,V$$
B.
$$4\,V$$
C.
zero
D.
$$6.4\,V$$
Answer :
$$4\,V$$
Solution :
$$\eqalign{ & {W_{{\text{el}}{\text{.}}}} = q\left( {{V_i} - {V_f}} \right) \cr & {\text{or}}\,\,6.4 \times {10^{ - 19}} = - 1.6 \times {10^{ - 19}}\left( {{V_A} - {V_B}} \right) \cr & {\text{or}}\,\,{V_A} - {V_B} = - 4\,V \cr & {\text{or}}\,\,{V_A} - {V_C} = - 4\,V\,\,\left( {\because {V_B} = {V_C}} \right) \cr & {\text{or}}\,\,{V_C} - {V_A} = 4\,V \cr} $$
$$\eqalign{ & {W_{{\text{el}}{\text{.}}}} = q\left( {{V_i} - {V_f}} \right) \cr & {\text{or}}\,\,6.4 \times {10^{ - 19}} = - 1.6 \times {10^{ - 19}}\left( {{V_A} - {V_B}} \right) \cr & {\text{or}}\,\,{V_A} - {V_B} = - 4\,V \cr & {\text{or}}\,\,{V_A} - {V_C} = - 4\,V\,\,\left( {\because {V_B} = {V_C}} \right) \cr & {\text{or}}\,\,{V_C} - {V_A} = 4\,V \cr} $$