Point charge q 1 = +7.95 C is at point A (-13.8 cm,0), and point charge q 2 , ca

Question

Point charge q1 = +7.95 C is at point A (-13.8 cm,0), and point charge q2, carrying an equal and opposite charge, is at point B (+13.8 cm,0).

a) Find the magnitude of the electric field at the origin.
E = 1 N/C

b) Give the direction of the electric field at the origin.

into the page zero     downward out of the page toward the negative charge toward the positive charge upward

c) An electron (q = 1.60x10-19, m = 9.11x10-31 kg) is placed at the origin and released. Find the magnitude of the acceleration the electron would experience, at the instant it is released, due to the two charges.
a = 3 m/s2

d) Give the direction of the force on the electron.

toward the positive charge zero     out of the page into the page downward upward toward the negative charge

Explanation / Answer

a.)

Electric field due to positive charge at origin = KQ/R^2 = 9x10^9 * 7.95*10^-6 / (0.138)^2 = 3.757 * 10^6 N/C

Electric field due to negative charge at origin =KQ/R^2 = 9x10^9 * 7.95*10^-6 / (0.138)^2 = 3.757 * 10^6

net electric field at origin = 7.514 * 10^ 6 N/C

b.)  toward positive axis, the charge at -ve x axis produce field towards +x axis and negative charge also produce electric field toward +x axis

c.) Force = Q * E = 1.6x 10^ -19 * 7.514* 10^6 = 1.202 * 10^-12 N

Force = mass * acceleration

a = 1.202 x 10^-12 / 9.1 * 10^-31

a = 1.3211 * 10^ 18 m/s^2

d.)direction of force is toward negative x -axis or toward positve charge as it is electron so it will move toward positive charge.

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