# electron in magnetic and electric field

I have this problem where i should find the direction and magnitude of the electric and the magnetic force on the electron. And then I'm supposed to find the direction and magnitude of the acceleration.
E = 1000 N/m B = 2,5 T v = 500 m/s

X |X X| X
¤->
X |X X| X
X |X X| X
V   V


I've found the electric force by Fe = E*q. With direction out of the screen by the right hand rule. Magnetic force = qvb in the y direction (meaning up) by the right hand rule. Now Im supposed to find the acceleration but I'm a bit stuck here. What i'm struggling with is to combine the two forces on the electron. Can someone get me in the right direction?

• Am I missing something? Don't you just add the two forces (vector addition) to get a total force $\mathbf F$ then use $\mathbf F = m \mathbf a$? – John Rennie Apr 27 '16 at 16:18
• So F = Fm + Fe? – KimR Apr 27 '16 at 16:20
• Using vector addition, yes. It isn't obvious from your diagram what the direction of $\mathbf E$ is. – John Rennie Apr 27 '16 at 16:22
• it's down. The answer made sense. Guess i was lost in my head a moment there. Thanks for the simple but valuable input :) – KimR Apr 27 '16 at 16:23

You just use vector addition and Newton's Second Law. For example, if you have $$\overrightarrow{F}_E=qE\hat{x},\space\space\space\overrightarrow{F}_B=qvB\hat{y}$$ then your total force $F_{tot}=F_E+F_B$ is just $$\overrightarrow{F}_{tot}=qE\hat{x}+qvB\hat{y}$$ Since $\hat{x}$ and $\hat{y}$ are totally linearly independent, these terms cannot be combined. Then using $F=ma$ you can deduce that $$\overrightarrow{a}=\frac{q}{m}(E\hat{x}+vB\hat{y})$$ but if you wanted the magnitude of the acceleration, you just take $$a=\sqrt{\overrightarrow{a}\cdot\overrightarrow{a}}=\frac{q}{m}\sqrt{E^2+v^2B^2}$$
• I'm not sure what you mean. The force exerted by an electric field is just $\overrightarrow{F}_E=q\overrightarrow{E}$, so if you know the field finding the force is trivial. – ocket8888 May 3 '16 at 7:55