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

$a = \frac{F}{m}$

$a = \frac{9.76 * 10^{-16}}{9.1 * 10^{-31}}$

$a = 1.07 * 10^{15} m/s^2$

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