In moving from A to B along an electric field | vedclass

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(B) The work done by the electric field on a charge $q$ in moving from point $A$ to point $B$ is given by $W = q(V_A - V_B)$.For an electron,$q = -e$,

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$4$

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(B) The work done by the electric field on a charge $q$ in moving from point $A$ to point $B$ is given by $W = q(V_A - V_B)$.For an electron,$q = -e$,where $e = 1.6 	imes 10^{-19} \, C$.Given $W = 6.4 	imes 10^{-19} \, J$ for moving from $A$ to $B$.Since $B$ and $C$ lie on the same equipotential surface $\phi_2$,we have $V_B = V_C$.Thus,$W = -e(V_A - V_B) = e(V_B - V_A) = e(V_C - V_A)$.Substituting the values: $6.4 	imes 10^{-19} = (1.6 	imes 10^{-19}) 	imes (V_C - V_A)$.Therefore,$(V_C - V_A) = \frac{6.4 	imes 10^{-19}}{1.6 	imes 10^{-19}} = 4 \, V$.

In moving from $A$ to $B$ along an electric field line,the electric field does $6.4 \times 10^{-19} \, J$ of work on an electron. If $\phi_1$ and $\phi_2$ are equipotential surfaces,then the potential difference $(V_C - V_A)$ is.....$V$.

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