Question

What are the magnitude and direction of the acceleration of an electron at a point where the electric field has magnitude 6100 n/c and is directed due north?

Answer 1

Force on electron due to electric field is given by

`F = eE`

`F = 1.6 * 10^(-19)* 6100`

`F = 9.76 * 10^(-16) N`

now the acceleration is given by

`a = (F)/(m)`

`a = (9.76 * 10^(-16))/(9.1 * 10^(-31))`

`a = 1.07 * 10^(15) m/s^2`

so above is the magnitude of acceleration and its direction is opposite to field as electron is negatively charged so direction is towards SOUTH