# Question cd7c0

Aug 15, 2017

Probably I got it wrong....but, $\frac{E}{m}$ represent the ratio between charge and mass of the electron.
The value of $\frac{E}{m}$ is the result of the famous experience carried out by J. J. Thomson to detect the electron.
E/m=1.7588×10^(11)C/(kg)#