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(B) The electric field component $E_y$ along the $Y$-axis is given by the negative potential gradient: $E_y = -\frac{\Delta V}{\Delta y}$.Given: $q =

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$24 	imes 10^{-4} \ N$

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(B) The electric field component $E_y$ along the $Y$-axis is given by the negative potential gradient: $E_y = -\frac{\Delta V}{\Delta y}$.Given: $q = 4 \ \mu C = 4 	imes 10^{-6} \ C$,$\Delta V = V_B - V_A = 54.8 \ V - 56 \ V = -1.2 \ V$,and $\Delta y = y_B - y_A = 12.5 \ cm - 12.3 \ cm = 0.2 \ cm = 0.2 	imes 10^{-2} \ m$.The electric field is $E_y = -\frac{-1.2}{0.2 	imes 10^{-2}} = \frac{1.2}{0.002} = 600 \ V/m$.The force component $F_y$ on the charge $q$ is $F_y = q E_y = (4 	imes 10^{-6} \ C) 	imes (600 \ V/m) = 2400 	imes 10^{-6} \ N = 24 	imes 10^{-4} \ N$.

Two points $A$ and $B$ on the $Y$-axis are at distances of $12.3 \ cm$ and $12.5 \ cm$ from the origin,respectively. The potentials at these points are $56 \ V$ and $54.8 \ V$,respectively. What is the component of the force on a charge of $4 \ \mu C$ placed at point $A$ on the $Y$-axis?

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