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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?

Respuesta :

Force on electron due to electric field is given by

[tex]F = eE[/tex]

[tex]F = 1.6 * 10^{-19}* 6100[/tex]

[tex]F = 9.76 * 10^{-16} N[/tex]

now the acceleration is given by

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{9.76 * 10^{-16}}{9.1 * 10^{-31}}[/tex]

[tex]a = 1.07 * 10^{15} m/s^2[/tex]

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

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