Consider a gravity-free container as shown. T | vedclass

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(D) The electric potential is $V = y^3 + 2$. The electric field is given by $\vec{E} = -\frac{dV}{dy} \hat{j} = -3y^2 \hat{j}$.At the base $(y=0)$, $V

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$76$

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(D) The electric potential is $V = y^3 + 2$. The electric field is given by $\vec{E} = -\frac{dV}{dy} \hat{j} = -3y^2 \hat{j}$.At the base $(y=0)$, $V_A = 2 	ext{ V}$. At the top face $(y=2 	ext{ m})$, $V_B = 2^3 + 2 = 10 	ext{ V}$.Using the work-energy theorem: $q(V_A - V_B) = \frac{1}{2}mv^2$. Since $q = -0.5 	ext{ C}$, we have $-0.5(2 - 10) = \frac{1}{2}(2)v^2$, which gives $4 = v^2$, so $v = 2 	ext{ m/s}$ (velocity just before collision).The velocity just before collision is $\vec{v}_i = 2 \hat{j} 	ext{ m/s}$. The velocity just after collision is $\vec{v}_f = -1.5 \hat{j} 	ext{ m/s}$.The change in momentum is $\Delta \vec{p} = m(\vec{v}_f - \vec{v}_i) = 2(-1.5 - 2)\hat{j} = -7 \hat{j} 	ext{ kg m/s}$.The average force exerted by the wall on the ball is $\vec{F}_{wall\_on\_ball} = \frac{\Delta \vec{p}}{\Delta t} = \frac{-7}{0.1} \hat{j} = -70 \hat{j} 	ext{ N}$.By Newton's third law, the force exerted by the ball on the wall is $\vec{F}_{ball\_on\_wall} = +70 \hat{j} 	ext{ N}$.Additionally, the electric field at the top face $(y=2)$ is $\vec{E} = -3(2)^2 \hat{j} = -12 \hat{j} 	ext{ N/C}$.The electric force on the ball is $\vec{F}_e = q\vec{E} = (-0.5)(-12)\hat{j} = 6 \hat{j} 	ext{ N}$.For the container to remain fixed, the external force $\vec{F}_{ext}$ must balance the force from the ball and the electric force on the ball: $\vec{F}_{ext} + \vec{F}_{ball\_on\_wall} + \vec{F}_e = 0$.$\vec{F}_{ext} + 70\hat{j} + 6\hat{j} = 0 \Rightarrow \vec{F}_{ext} = -76 \hat{j} 	ext{ N}$.The magnitude of the required external force is $76 	ext{ N}$.

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