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(D) The electron beam is moving out of the plane of the paper (towards the observer). Let the velocity be $\vec{v} = v \hat{k}$.The electric field $\v

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(D) The electron beam is moving out of the plane of the paper (towards the observer). Let the velocity be $\vec{v} = v \hat{k}$.The electric field $\vec{E}$ is directed downwards,so $\vec{E} = -E \hat{j}$. The force on the electron due to the electric field is $\vec{F}_e = -e \vec{E} = eE \hat{j}$. This causes a deflection in the positive $y$-direction.The magnetic field $\vec{B}$ is directed downwards,so $\vec{B} = -B \hat{j}$. The force on the electron due to the magnetic field is $\vec{F}_m = -e (\vec{v} 	imes \vec{B}) = -e (v \hat{k} 	imes -B \hat{j}) = -e (vB \hat{i}) = -evB \hat{i}$. This causes a deflection in the negative $x$-direction.Since the magnitudes of the deflections are equal,the net force is $\vec{F}_{net} = \vec{F}_e + \vec{F}_m = eE \hat{j} - evB \hat{i}$.The spot will be deflected in the second quadrant (negative $x$,positive $y$).Comparing this with the given options,diagram $D$ shows the spot in the second quadrant.

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