1.     A charge of +3.5 µC is placed on the x axis at x = +0.55 m, while a charge of -15 µC is placed at the origin.

(a) Calculate the magnitude and direction of the net electric field on the x-axis at x = +0.8 m.

(b) Determine the magnitude and direction of the force that would act on a charge of – 8.0 µC if it was placed on the x axis at x = + 0.8 m.

 

Solution:

a)     The electric field from a point charge is given by E = k|q|/r^2

                        E points towards a – charge and array from a + charge.

                        Both electric field points in the same direction so we just add them together

                        E (net) =

(8.99*10^9 Nm^2/C^2 * 3.5*10^-6 C)/ 0.55^2 + (8.99*10^9 Nm^2/C^2 * 15*10^-6 C)/ 0.8^2

                        E (net) = 2.9* 10^5 N/C +x direction

b) ?

 

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